## Summary of Metric Space Relations - Continuous map $f:X \to Y$ preserve: - Openness of $B \subseteq Y$ when taking the preimage $f^{-1}(B)$. - Connectness and path connectedness of $X$. - Sequential compactness and compactness of $X$. - Homeomorphism $\phi:X \to Y$ preserve: - Open/closedness of any $A \subseteq X$. - Anything preserved by continuous maps. - Open/closedness of a set is preserved under homeomorphisms. - Openness is also preserved by taking the preimage of a continuous map. - Switching between equivalent metrics preserve: - Open/closedness of sets. - Continuity. - Switching between Lipschitz/strongly equivalent metrics preserve: - Completeness. - Anything preserved by equivalent metrics. - "Uniform homeomorphisms" (i.e. homeomorphisms that are uniformly continuous) preserve completeness. ## The Metric Function * A **metric space** is a set $X$ equipped with a **metric** (or distance function) $d: X\times X \to \mathbb{R}$ that satisfies the following axioms: * Positive-definiteness: $d(x,y)\ge 0$, with equality only when $x=y$. * Symmetry: $d(x,y)=d(y,x)$. * Triangular inequality: $d(x,y)+d(y,z)\ge d(x,z)$. * If the set $X$ is also a vector space, then a function $\|\cdot\|:X \to \mathbb{R}$ is called a **norm** if it satisfies: * Positive-definiteness: $\|x\|\ge 0$, with equality only when $x=0$. * Linear scaling: $\|\lambda x\|=|\lambda|\|x\|$ * Triangle inequality: $\|x+y\| \le\|x\|+\|y\|$ * Given a norm on $X$, it **induces** the metric $d(x,y)=\|x-y\|$. ### Common Metrics and Norms * In $\mathbb{R}^n$, the $l_{1},l_{2},l_{\infty}$ norms of $\mathbf{x}=(x_{i})_{i=1}^{n}$ are given by$\begin{align*} &l_{1}(\mathbf{x})=\sum_{i}|x_{i}|\\ &l_{2}(\mathbf{x})=\sqrt{ \sum_{i} x_{i}^2 }\\ &l_{\infty}(\mathbf{\mathbf{x}})=\sup_{i}|x_{i}| \end{align*}$and the metrics they induce are called $d_{1},d_{2},d_{\infty}$. Note that $d_{2}$ is just the usual Euclidean distance. * In any set, the **discrete metric** $d(x,y)=\mathbb{1}_{x\ne y}$ is a metric. It's often a good counterexample to seemingly legit statements. * In the vector space of functions bounded on $X$, denoted $B(X)$, the **supremum norm** induces the **supremum metric**: $\begin{align*} \|f\|_{\infty}&= \sup\{ f(x): x \in X \}\\\\ d_{\infty}(f,g)&= \|f-g\|_{\infty} \end{align*}$ * The set of continuous functions on an interval $A$, denoted $C(A)$, can be equipped with the $d_{1}$ metric $d_{1}(f,g)=\int _{A} |f-g| \, dx$ * Given metric spaces $X,Y$, the set $X\times Y$ forms a metric space with the **product metric**: $d_{X\times Y}=\sqrt{d_X^2+d_Y^2}$ * For example, $(\mathbb{R}^{2},d_{2})=\mathbb{R}\times \mathbb{R}$. ## Open and Closed Sets ### Disks * The **(open/closed) ball** in the metric space $X$ around a point $x_{0}$ of radius $r$ is $\begin{align*} \text{open}: \,&B(x_{0}, r)= \{ x\in X:d(x,x_{0})<r \}\\ \text{closed}: \,&\bar{B}(x_{0},r) = \{ x \in X: d(x,x_{0})\le r \} \end{align*}$and if necessary, use superscript to indicate the metric used: $B^{d_{X}}(x_{0},r)$. * Note that given different metrics, this set is not necessarily "round", and it might be things like: * A diamond in $(\mathbb{R}^2,d_{1})$ and a square in $(\mathbb{R}^{2},d_{\infty})$ * A single point ($r<1$) or the entire set ($r \ge 1$) under the discrete metric * A set of functions in $(B(X),d_{\infty})$ ### Open and Closed Sets * A subset $A \subseteq X$ in the metric space is an **open set** when: $\forall a \in A,\, \exists r>0: B(a,r) \subseteq A$ * *That is, there is some elbow-room around each point* (quoted from Sutherland). * By definition, the empty set and the metric space itself are open (hence it is sometimes important to specify the metric space: $[a,b]$ is open in itself, but not in $\mathbb{R}$). * A subset is a **closed set** when its complement is open. * <u>Closed sets are not necessarily "not open"</u>: by definition, the empty set and the entire space are both closed and open. * Sets can also be neither closed or open, for example $[0,1) \subset \mathbb{R}$. ### Basic Results * An open ball $B(x, r) \subseteq X$ is an open set. > Proof: take any point $y \in B(x,y)$, and write $l=d(x,y)<r$. > Then consider the open ball $B^*=B\left(y, \frac{(r-l)}{2} \right)$. Any point $z \in B^*$ must be within a distance of $r$ from $x$ due to the triangle inequality. > Hence $\exists \delta=\frac{r-l}{2}: B(y,\delta) \subseteq B(x,r)$, and $B(x,r)$ is an open set. - A closed ball $\bar{B}(x,r)$ is a closed set. > Proof: any $y \in \bar{B}(x,r)^{c}$ has $d(x,y)=\rho>r$. Then use triangular inequality to prove that $B\left( y,\frac{\rho-r}{2} \right) \subseteq \bar{B}(x,r)^{c}$. * A finite intersection of open sets is also open. * Could fail for an infinite intersection: $\{ 0 \}=\cap_{n\in \mathbb{N}}\left( -\frac{1}{n}, \frac{1}{n} \right)$ is closed in $\mathbb{R}$. > Proof: assume $A_{i}$ all open for $i \in \{ 1, \dots, n \}$, and take any $x \in\cap_{i} A_{i}$. > The definition of open sets gives $r_{i}:B(x,r_{i}) \subseteq A_{i}$, and take their minimum $r=\min\{ r_{i} \}$ (positive since the intersection is finite). Then verify that $B(x,r) \subseteq \cap_{i} A_{i}$. * An arbitrary (finite or infinite) union of open sets is also open. * Hence in $\mathbb{R}$, $(a,\infty)=\cup_{n\in \mathbb{N}}\left( a+\frac{1}{n},a+n \right)$ is open, and $(-\infty,a]=(a,\infty)^c$ is closed. > Proof: assume $A_{i}$ are all open, and take any $x \in \cup_{i} A$, say $x\in A_{n}$. Apply the definition of openness on $x\in A_{n}$ gives $B(x,r) \subseteq A_{n} \subseteq \cup_{i} A_{i}$. * A finite union of closed sets is also closed; an arbitrary intersection of closed sets is also closed. > Proof: apply the previous results to their complements. ## Limit Points, Closures, and Interiors ### Limit and Isolated points * A **limit point** of a set $Y$ in a metric space $X$ is a point $x \in X$ such that: $\forall \epsilon>0, \ B(x,\epsilon)\cap \big(Y\backslash\{x\}\big)\ne\emptyset$ * *That is, $x$ is touching $Y$*, though the <u>limit point does not have to be in the set</u>. * E.g. both $0$ and $1$ are limit points of $[0,1)$. * The set of limit points of $Y$ is denoted $L(Y)$. * An **isolated point** of $Y$ is a point $y\in Y$ such that: $\exists\epsilon>0:\ B(y,\epsilon)\cap Y=\{y\}$ * *That is, nothing else is close to $y$ in $Y$*. * The union of the limit points and isolated points of $Y$ is called its **points of closure**. * Note that any point in $Y$ is either a limit or an isolated point. Hence $Y$ is a subset of its points of closure. * A point is a limit point iff it can be approached by a sequence: $\exists(x_{n})\subseteq X,(x_{n})\to x \iff x \in L(X)$ > Proof: easy result from the definitions by taking $\epsilon= \frac{1}{ n}$, for example. * A set $Y$ is closed $\iff$ it contains all its limit points. > Proof: ($\Rightarrow$) if for contradiction a limit point $x$ is not in $Y$ closed, then > by openness of $Y^c$, there is $r: B(x,r) \subseteq Y^c$. > Then the definition of a limit point gives $y \in Y: y\in B(x,r) \subseteq Y^c$, a contradiction. > > ($\Leftarrow$) we shall show that $Y^c$ is open. For any $a \in Y^c$, if for contradiction $\forall \epsilon>0, \exists y \in Y\cap B(a,\epsilon)$, then $a$ is a limit point of $Y$, which by assumption is in $Y$, a contradiction. So $B(a,\epsilon)\subseteq Y^c$, $Y^c$ is open, $Y$ is closed. * The set of limit points $L(Y)$ of subset $Y \subseteq X$ is closed. > Proof: ![[LimitPointsClosed.png]]We will prove that $L(Y)^c$ is open by finding an open ball around any $a \in L(Y)^{c}$. > Since $a\not\in L(Y)$, there $\exists r>0: B(a,r) \cap Y=\emptyset$ or $\{ a \}$. However, this is not enough as $L(Y) \not \subseteq Y$ in general. We shall show instead that $B(a, r / 2)$ is the desired open ball. > > Fix any $b \in B(a, r/2)$, and if $b=a$, then certainly $b \in L(Y)^c$. > If $b \ne a$ (see diagram), then $B(b,d(a,b) / 2) \cap Y=\emptyset$, so $b\in L(Y)^c$. > Hence $B(a,r / 2)\subseteq L(Y)^c$, and the set is open. ### Closures * Given a set $Y \subseteq X$, its **closure** is the intersection of all closed sets that contain $Y$: $\bar{Y},\text{Cl}(Y),\text{Clo}(Y)\equiv\bigcap_{\substack{Y\subseteq U\\ U\text{ closed}} }U$ * Closures are different in different metric spaces, so be explicit if necessary: in $(0,2)$, $[1,2)$ is its own closure, but in $\mathbb{R}$, its closure is $[1,2]$. * Equivalent definition: $\bar{Y}$ is the smallest closed set that contains $Y$. > Proof of equivalence: the closure defined above is an infinite intersection of closed sets, which is also closed. * A subset $Y \subset X$ is **dense** if $\bar{Y}=X$. * *That is, every point in $X$ is touching $Y$.* * For example, $\mathbb{Q}$ is dense in $\mathbb{R}$. * The closure equals the points of closure (the limit and isolated points), hence <u>the closure of a set is its union with its limit points</u>: $\bar{Y}=Y \cup L(Y)$ > Proof: take $Y\subseteq X$, and let $\bar{Y}$ be its closure, and $L(Y)$ be its limit points. > ($\bar{Y}\subseteq Y \cup L(Y)$) If for contradiction $y \in \bar{Y}$ is neither a limit or an isolated point, then there is a radius $r:B(y,r) \cap Y=\emptyset$. Removing this ball from $\bar{Y}$ gives a smaller closed set that contains $Y$, a contradiction to the (equivalent) definition of closures. > > ($\bar{Y} \supseteq Y \cup L(Y)$) Obviously isolated points, being within the set, are in the closure. Limit points of $Y$ are in the closure since they are also limit points of $\bar{Y}$, and the closed set $\bar{Y}$ contains all its limit points. ### Interiors and Boundaries * The **interior** of $Y$ is the union of all open sets contained in $Y$: $\text{int}(Y),\, \mathring{Y} \equiv\bigcup_{\substack{U\text{ open}\\U\subset Y}}U$ * For example, $\mathrm{int} (\mathbb{Q})=\emptyset,\, \text{int}{[0,1)}=(0,1)$ in $\mathbb{R}$. * Equivalent definition: $\mathring{Y}$ is the largest open subset of $Y$. * Hence $(\mathring{Y})^c$ is the smallest closed set containing $Y^c$, aka its closure $\overline{Y^c}$. * Another equivalent definition: $\mathring{Y}=\{ y \in Y \,|\, \exists \epsilon>0: B(y,\epsilon) \subseteq Y \}$. * The **boundary** $\partial Y$ of $Y$ is $\bar{Y}\setminus \mathring{Y}$. * For example, in $\mathbb{R}$, $\partial [0,1]=\{ 0,1 \}$, and $\partial \mathbb{Q}=\mathbb{R}$. ## Continuity and Convergence * A map $f:X \to Y$, where $X,Y$ are metric spaces with metrics $d_{X},d_{Y}$ respectively, is **continuous** at $x_{0}$ if: $\forall \epsilon>0,\, \exists \delta>0: \forall x\in X,\, [d_{X}(x,x_{0})<\delta] \Rightarrow [d_{Y}(f(x),f(x_{0}))< \epsilon]$which is the generalization of continuity in $(\mathbb{R},l_{2})$ learned in Prelims. To remove ambiguity, denote this as $(d_{X},d_{Y})$-continuity. * Equivalently, the definition can be formulated in terms of disks: $\forall \epsilon>0,\, \exists \delta>0: f(B^{d_{X}}(x_{0},\delta)) \subseteq B^{d_{Y}}(f(x_{0}), \epsilon)$ * In a metric space $(X,d)$, a sequence $(x_{n})$ **converges** to $x\in X$ if$\forall \epsilon>0,\, \exists N \in \mathbb{N}: \forall n \ge N,\, x_{n}\in B^d(x,\epsilon)$ * Note that <u>convergence requires the limit to be within the metric space</u>: $\left( \frac{1}{n} \right) \subset [0,1]$ converges to $0$, but fails to converge in $(0,1]$. * A sequence $(x_{n})$ is **Cauchy** in a metric space if $\forall \epsilon>0, \exists N \in \mathbb{N}: \forall m,n \ge N, d(x_{m},x_{n}) < \epsilon$ ### Boundedness * A subset $A \subseteq X$ in a metric space is **bounded** if all of it is within a finite distance of a point: $\exists x_{0} \in X, r\in \mathbb{R}: A\subseteq B(x_{0}, r)$ * The **diameter** of a bounded set $A$ is $\mathrm{diam}(A)\equiv \sup_{x,y\in A}d(x,y)$. * A function $f: X \to Y$ is bounded over $A \subseteq X$ if $f(A)$ is bounded in $Y$. * Finite unions of bounded sets are bounded. > Proof: say there are bounded subsets $A_{1}, \dots, A_{n}\subseteq X$, and the definition of boundedness gives $x_{i},r_{i}:A_{i} \subseteq B(x_{i},r_{i})$. > Take any $x_{0}\in X$, and we shall prove that all points of the union are within a certain radius from it. > Write $d_{i}=d(x_{0},x_{i})$, then triangular inequality requires $\forall a\in A_{i}, d(a,x_{0})\le d(a,x_{i})+d(x_{0},x_{i})\le r_{i}+d_{i}$. > Hence taking $\max_{i} \{ r_{i} +d_{i} \}$ gives the required radius. ### Basic Results * The limit is unique, just like in $\mathbb{R}$. > Proof: take two limits $x,y$, and write their distance $l=d(x,y)$. > Applying the definition with $\epsilon=l / 2$ gives a contradiction unless $l=0$. Then positive-definiteness of the metric gives $x=y$. * All convergent sequences are Cauchy. The converse is not generally true, unlike in $\mathbb{R}$. > Proof: similar to the proof for the uniqueness of the limit. Show that terms are at most $\epsilon / 2$ around the limit, and use the triangle inequality. > > Counterexample to the converse: $\left( \frac{1}{n} \right) \subset (0,1]$ is Cauchy but does not have a limit in the space. * A sequence $((x_n,y_n))\subseteq X\times Y$ converges $\iff$ the two component sequences $(x_n),(y_n)$ converge in $X,Y$. > ($\Rightarrow$) by sandwiching $d_X(x_n,x),d_Y(y_n,y)$ between $d_{X\times Y}$ and $0$. > ($\Leftarrow$) by AOL. * As in $(\mathbb{R},l_{2})$, the sum, modulus, and product of continuous functions are still continuous. And if the denominator is non-zero in the neighborhood, so are the quotients. * If $f:X\to Y$ is continuous at $x$, and $g:Y\to Z$ is continuous at $f(x)$, then their composition $f \circ g: X \to Z$ is also continuous at $x$. * If $f:X\to Y$ is continuous at $x$, and $g:U \to V$ is continuous at $u$, then their "Cartesian product" $(f \times g):(X \times U)\to(Y \times V)$ is also continuous at $(x,u)$. ### Open/Closedness and Continuity * A map $f: X \to Y$ is continuous $\iff$an open subset $V \subseteq Y$ always has an open preimage $f^{-1}(V) \subseteq X$. * <u>Continuous functions do not map open sets to open sets in general</u>: take the zero function for example. > Proof:![[ContinuityOpenPreimage.png]] > ($\Rightarrow$) Assume $f$ continuous and $V \subseteq Y$ open. For any point $u \in f^{-1}(V)$, openness of $V$ gives $\epsilon>0:B(f(u),\epsilon) \subseteq V$. > Continuity of $f$ then gives $\delta>0: f(B(u, \delta)) \subseteq B(f(u),\epsilon)$. So $B(u, \delta)$ is an open ball in the preimage, and hence the preimage is open. > > ($\Leftarrow$) Any $B(f(u), \epsilon)$ is open, hence by assumption it has an open preimage. Hence there is a ball $B(u, \delta) \subseteq$ the preimage, satisfying the disk definition of continuity. * Similarly, $f$ continuous $\iff$ closed subsets have closed preimages. > Proof: apply the previous result on their complements. ## Homeomorphisms and Equivalent Metrics ### Homeomorphisms * A **homeomorphism** between metric spaces $X,Y$ is a bijection $f:X \to Y$ such that both $f$ and $f^{-1}$ are continuous. * <u>Homeomorphisms are equivalence relations</u>, as compositions of homeomorphisms are also homeomorphisms. * $\mathbb{R},(a,b),(a,\infty),(-\infty,a)$ are all homeomorphic: * $(-1,1) \to \mathbb{R}$ is given by $f(x)=\frac{x}{1-|x|}$. * $(0, \infty)$ is homeomorphic with $(0,1)$ by the map $\frac{1}{x}$. Similarly for $(-\infty, a)$ via $-\frac{1}{x}$. * $(-1,1)$ and $(0,1)$ are homeomorphic with other finite open intervals by shifting and scaling. > Proof (for $f$): $f$ is continuous since the absolute value, quotient, and products are all continuous in the respective intervals. The inverse is continuous since $f$ continuous and monotone on any interval (continuous inverse function theorem from Prelims Analysis II). * <u>Homeomorphisms are open mappings</u> (i.e. they map open sets to open sets) * Hence closed sets like $[-1,1]$ are not homeomorphic to $\mathbb{R}$. > Proof: use the fact that under continuous maps, open sets have open preimages, and apply it to both $f$ and $f^{-1}$. ### Equivalent Metrics * Two metrics $d, \tilde{d}$ on $X$ are **equivalent** if $d$-openness $\iff$ $\tilde{d}$-openness. * For example, $d_{1,2,\infty}$ are all equivalent in $\mathbb{R}^n$, but not with the discrete metric. * The $d_{1}$ (integral) and $d_{\infty}$ (supremum) metrics on $C[a,b]$ are not equivalent: there is no $B^{d_{1}}(0,\epsilon) \subseteq B^{d_{\infty}}(0,1)$ for any $\epsilon >0$. * Equivalent definition: two metrics are equivalent iff the identity map $(X,d) \to (X, \tilde{d})$ is a homeomorphism. > Proof: ($\Rightarrow$) bijection is immediate. Then use the fact that "open sets have open preimages" implies continuity for both $f$ and $f^{-1}$ , and apply it to the identity map here. > > ($\Leftarrow$) immediate from the fact that homeomorphisms are open mappings. * <u>Continuity is shared in equivalent metrics</u>: if $d, \tilde{d}$ are equivalent in $X$, then for maps $X \to (Y, d_{Y})$, continuity under $(d,d_{Y})$ $\iff$ continuity under $(\tilde{d},d_{Y})$. > Proof: immediate from the equivalence between continuity and "open sets have open pre-images". * Two metrics $d_{1}, d_{2}$ n $X$ are **Lipschitz/strongly equivalent** if: $\exists r,R:\, \forall x,y \in X,\,rd_{2}(x,y) \le d_{1}(x,y) \le Rd_{2}(x,y)$ * Note that $d_{1},d_{2},d_\infty$ in $\mathbb{R}^{n}$ are Libschitz equivalent. * Lipschitz equivalence implies equivalence. > Proof: we show that $d_{1}$-openness implies $d_{2}$ openness. The other way around is symmetric. > > Lipschitz equivalence guarantees that any $d_{1}$-open ball $B^{d_{1}}(x,\delta)$ contains a $d_{2}$-open ball $B^{d_{2}}(x,\delta / r)$. So if $X$ is $d_{1}$-open, any $x \in X$ has a $d_{2}$-open ball around it, hence $X$ is also $d_{2}$-open. ## Completeness * A metric space $X$ is **complete** if all Cauchy sequences in the space converges. * For example, $\mathbb{R}$ is complete, but $\mathbb{Q}$ and $(0,1)$ are not. Hence <u>completeness is not preserved under homeomorphisms</u>, as $\mathbb{R}$ is homeomorphic to $(0,1)$. ### Criteria for Completeness * <u>Closedness equals completeness</u>: if $Y \subseteq X$, $X$ complete, then $Y$ complete $\iff$ $Y$ is closed in $X$. > Proof: ($\Rightarrow$) all $y \in L(Y)$ have a sequence $(y_{n})\subseteq Y:(y_{n})\to y$. $(y_{n})$ must be Cauchy since it converges, and so $Y$ complete means $y\in Y$. So $L(Y)\subseteq Y$, $Y$ closed. > > ($\Leftarrow$) take Cauchy sequence $(y_{n})\subseteq Y\subseteq X$, and since $X$ complete, $\exists y:(y_{n})\to y$, and $y$ must be a limit point of $Y$. Since $Y$ closed, $y\in Y$, and hence $Y$ complete. * A product space is complete $\iff$ all its components are complete. * As a corollary, $\mathbb{R}^n$ is complete since $\mathbb{R}$ is complete. > Hint: although the usual product space is defined with the "$d_{2}quot; distance, that is $d_{\text{prod}}=\sqrt{d_{X}^2+d_{Y}^2}$, by Lipschitz equivalence of $d_{1},d_{2},d_{\infty}$, we can prove the result for $d_{\text{prod}}=\max(d_{X},d_{Y})$ instead. > Then just work out the details for Cauchy-ness and convergence. * Completeness is perserved by a "uniform homeomorphism", a bijection $f: X \to Y$ where $f, f^{-1}$ are both uniformly continuous. * As a result, completeness is preserved in switching between Lipschitz/strongly equivalent metrics (consider the switch as the identity map $(X,d) \to (X,\tilde{d})$) > Proof: say $f$ is a uniform homeomorphism, with $d(x_{1},x_{2})<\delta \Rightarrow d(f(x_{1}),f(x_{2}))<\epsilon$, the same $\delta$ valid for all $x_{1},x_{2}$. We shall show that (1) $f$ maps Cauchy sequences to Cauchy sequences, and (2) preserves its convergence. > > <u>Cauchy-to-Cauchy</u>: given Cauchy sequence $(x_{n})$, taking $m,n$ large such that $d(x_{m},x_{n})<\delta$ means that the image-sequence $d(f(x_{m}),f(x_{n}))< \epsilon$, giving Cauchiness. > > <u>Convergence</u>: Given $\epsilon>0$, uniform continuity gives $\delta$. Since $X$ is complete, the Cauchy sequence $(x_{n}) \to x \in X$. Take $k$ large so that $d(x_{k}, x)<\delta$, so $d(f(x_{k}),f(x))<\epsilon$. So the Cauchy sequence $(f(x_{n}))$ converges to $f(x)\in Y$ (note that $f$ being bijection guarantees inclusion). * Cantor's intersection theorem: in a complete metric space $X$, $S_{1} \supseteq S_{2} \supseteq S_{3} \dots,\,\mathrm{diam}(S_{n}) \to 0$, $S_{i}$ are all closed and non-empty, then $\cap_{i}S_{i}=\{ a \}$ for some $a \in X$. > Proof: > <u>Finding the point</u>: take $x_{n} \in S_{n}$, then the sequence $(x_{n})$ is Cauchy. For any $\epsilon>0$, find $k:\mathrm{diam(S_{k})}<\epsilon$. Then when $m,n \ge k,\,x_{m,n} \in S_{k}$, so $d(x_{m},x_{n})< \mathrm{diam}(S_{k}) <\epsilon$. Completeness of $X$ guarantees convergence $(x_{n}) \to a \in X$. > > <u>Non-emptiness</u>: since the tail $(a_{n \ge k}) \subseteq S_{k}$, $a$ is a limit point of $S_{k}$, so $S_{k}$ closed implies $a \in S_{k}$. Since this is true for all $k$, $a \in \cap_{i} S_{i}$. > > <u>Uniqueness</u>: suppose $a,b \in \cap_{i} S_{i}$. Then $\forall k:a,b \in S_{k}$, and hence $d(a,b)<\mathrm{diam}(S_{k}) \to 0$, so $a=b$. ### Completeness of Function Spaces * $\big(B(X),d_{\infty}\big)$ is complete, where $B(X)$ is the set of bounded functions $X\to \mathbb{R}$. > Proof: take a Cauchy sequence $(f_{n})\subseteq B(X)$. > <u>The limit</u>: $(f_{n})$ Cauchy means that at each point $x\in X$, $(f_{n}(x))\subseteq \mathbb{R}$ is a Cauchy sequence. So by completeness of $\mathbb{R}$, the sequence converges pointwise to $f(x)\equiv \lim_{ n \to \infty }f_{n}(x)$. > > <u>Boundedness</u>, so that $f \in B(X)$: > Cauchy-ness of $(f_{n})$ in $d_{\infty}$ gives $N:\forall n\ge N,\, \sup_{x \in X}|f_{N}(x)-f_{n}(x)|<1$. > Then at any $x_{0}\in X$, $|f_{N}(x_{0})-f_{n}(x_{0})|<1$. > So AOL gives $|f_{N}(x_{0})-f(x_{0})|\le 1$, so $f(x_{0}) \le f_{N}(x_{0})+1$. Since $f_{N}$ is bounded, so is $f$. > > <u>Convergence</u> in $d_{\infty}$: Cauchy-ness gives $N:\forall m,n\ge N,\, |f_{m}(x)-f_{n}(x)|<\epsilon$, the same $N$ applicable at $\forall x \in X$. > so letting $m \to \infty$ gives $N:\forall n\ge N,x \in X,\, |f(x)-f_{n}(x)|\le\epsilon$, which is by definition $(f_{n}) \xrightarrow{d_{\infty}} f$. * The metric space of continuous functions over a metric space $(X,d) \to \mathbb{R}$, $(C(X),d_{\infty})$ is complete. > Proof: take a Cauchy sequence $(f_{n}) \subseteq C(X)$. Since $B(X)$ is complete, it has a limit $(f_{n}) \to f \in B(X)$. Furthermore, $f \in C(X)$ because convergence under the $d_{\infty}$ metric is in fact uniform convergence, hence continuity of $f_{n}$ implies continuity of $f$. ### Contraction Mapping Theorem - A map $f:(X,d_{X}) \to (Y,d_{Y})$ is **Lipschitz continuous** if all distances are scaled by at most a factor of $K \in \mathbb{R}^{+}$: for all $x, \tilde{x} \in X$, $Kd_{X}(x, \tilde{x}) \ge d_{Y}(f(x), f(\tilde{x}))$ - A map $f:(X,d_{X}) \to (Y,d_{Y})$ is a **contraction (mapping)** if it is Lipschitz continuous, and its factor $K \in [0,1)$: for all $x, \tilde{x} \in X$, $Kd_{X}(x, \tilde{x}) \ge d_{Y}(f(x), f(\tilde{x}))$that is, the distance is shrunk by at least a factor of $K$. - Lipschitz continuity implies (regular) continuity; in particular, contractions are continuous. > Proof: use $\delta-\epsilon$ definition of continuity, and let $\delta=\epsilon/ K$. - The **contraction mapping theorem (CMT)** states that in $(X,d)$ compact and non-empty, if $f:X \to X$ is a contraction, then there is a unique **fixed point** $x:f(x)=x$. > Proof: let $f$ have contraction constant $K$. > <u>Uniqueness</u>: if $x, \tilde{x}$ > <u>Existence</u>: start at any $x_{0} \in X$, define $(x_{n})$ by repeatedly applying $f$: let $x_{n}=f(x_{n-1})$. Then $\begin{align*} d(x_{n},x_{n+1}) &\le Kd(x_{n-1},x_{n}) \\ &\le \dots \\ &\le K^{n}d(x_{0},x_{1}) \end{align*}$Then for any $x_{m},x_{n}$ (WLOG let $m \ge n$), the triangular inequality gives $\begin{align*} d(x_{n},x_{m}) &\le d(x_{n},x_{n+1})+ \dots + d(x_{m-1},x_{m})\\ &\le (K^{n}+K^{n-1}+\dots+K^{m})\cdot d(x_{0},x_{1})\\ &\le \frac{K^{n}}{1-K}d(x_{0},x_{1}) \end{align*}$So by requiring $m,n \ge N$ for some large $N$, the distance $d(x_{m},x_{n})$ can be arbitrarily small, and $(x_{n})$ is Cauchy. Since $X$ is complete, $(x_{n}) \to x \in X$. Now $x$ is a fixed point since $f(x)=\lim_{ n \to \infty } f(x_{n})=\lim_{ n \to \infty } x_{n+1} =x$by continuity of $f$ as a contraction. ## Connectedness and Path Connectedness * A metric space $X$ is **disconnected** if it can be partitioned by two open, non-empty subsets of it, that is: $\exists A,B\text{ open} \subsetneq X:\, A \cap B= \emptyset,\, A \cup B = X,\,$and otherwise it is **connected**. * $A,B$ only need to be open in $X$, not some larger metric space (if relevant): $[1,2]$ and $[3,4]$ disconnect $[1,2] \cup [3,4] \subset \mathbb{R}$ even though they are not open in $\mathbb{R}$. * Equivalent definition: $X$ is connected iff any continuous map $f: X \to (\{ 0,1 \}, d_\text{discrete})$ must be constant. > Proof: ($\Rightarrow$) $\{ 0 \}$, being open under the discrete metric, has an open pre-image under $f$. So does $\{ 1 \}$. But then their pre-image would form a partition of $X$ by open sets, a contradiction unless one of the pre-images is empty. Hence $f$ constant. > > ($\Leftarrow$) If for contradiction $X$ is disconnected by $A,B$, then the map $f=\mathbb{1}_{x \in A}$ is continuous but not constant, a contradiction. * Another equivalent definition: $X$ is connected iff the only subsets of $X$ that are both open and closed are $X$ and $\emptyset$. ### Criteria for Connectedness * <u>Connectness is preserved by continuous maps</u>: under a continuous map, the image of a connected space is also connected. * Hence in particular, <u>connectedness is shared by homeomorphic spaces.</u> > Proof: suppose that $f: X \to Y$, $X$ connected, and for contradiction $f(X)$ is disconnected by $A, B \subsetneq f(X)$, $A,B$ both open and non-empty. > Then $f^{-1}(A)$ and $f^{-1}(B)$ are non-empty, open (being preimages of open sets), and disconnect $X$, a contradiction. * If $A$ is connected, so is any $B:A \subseteq B \subseteq \bar{A}$. * In particular, if a set is connected, so is its closure. > Proof: we show that any continuous function $f:B \to \{ 0,1 \}$ must be constant. Note that $f|_{A}$ is still continuous, so connectedness of $A$ requires $f|_{A}$ constant. > Then for any $b \in B$, either $b \in A$ and the result is immediate, or $b \in L(A)\setminus A$. So there is $(a_{n}) \subseteq A: (a_{n}) \to b$. Continuity of $f$ requires $(f(a_{n})) \to f(b)$, so $f(b)=f(A)$. Hence $f(B)=f(A)$ constant. * The sunflower lemma: in a metric space $X$, given a collection of connected subsets $\{ A_{i} \} \subseteq \mathcal{P}(X)$ satisfying $A_{i}\cap A_{j}\ne \emptyset$, their union $\cup_{i} A_{i}$ is also connected. * *This version is stronger than the lemma in the notes, but is easier(?) to prove. It doesn't look like a sunflower, though.* * *This name "sunflower lemma" is not used anywhere other than in the lecture notes, so I'm not sure if it's safe to quote by name only.* > Proof: we will show that any continuous function $f: \cup_{i}A_{i} \to \{ 0,1 \}$ must be constant. Define $b_{i}=f(x_{i})$ for some $x_{i} \in A_{i}$, well-defined since $f|_{A_{i}}$ is constant due to continuity and connectness. > > Then for any $i,j$, there is $x_{ij}\in A_{i} \cap A_{j} \ne \emptyset$, so $b_{i}=f(x_{ij})=b_{j}$. So $f$ must be constant. ### Connectedness in $\mathbb{R}^{n}$ *Big idea: interval equals connectedness.* * A set $S \subseteq \mathbb{R}$ has the **interval property** if $x,y \in S$, then $[x,y] \subseteq S$. * *That is, if two points are in the set, so are anything between them.* * In $\mathbb{R}$, a set is an interval $\iff$ it has the interval property. > Proof: ($\Rightarrow$) immediate. > ($\Leftarrow$) Work with $a=\inf S,b=\sup S$, replacing them with $\pm \infty$ if necessary, and show that $(a,b) \subseteq S \subseteq [a,b]$. * $S \subseteq \mathbb{R}$ is connected $\iff$ $S$ is an interval. * Hence $\mathbb{Q}$ is not connected. > Proof: ($\Rightarrow$) if for contradiction the connected set $S$ fails the interval property, there is $x,y \in S, z\in (x,y): z\not \in S$. Then verify that $S_{-}=\{ s\in S: s<z \}$ and $S_{+}= \{ s \in S: s>z \}$ disconnect $S$. > > ($\Leftarrow$) Note that if $(a,b)$ is connected, so is $[a,b]=\text{Closure}((a,b))$ ($a,b$ can be $\pm \infty$ in this case). But $(a,b)$ is homeomorphic to $\mathbb{R}$, which is connected. Since homeomorphisms preserve connectedness, all intervals are connected. * Continuous maps $f: X \to \mathbb{R}$ map connected spaces to intervals; in particular, the intermediate value theorem holds. > Proof: $f$ continuous must map connected spaces $X$ to connected subspaces in $\mathbb{R}$, which must be an interval. Any intermediate value would have a non-empty preimage, then. ### Path-Connectedness * A **path** in a metric space $X$ is a continuous map $p: [0,1] \to X$ * If we write $x=p(0)$ and $y=p(1)$, then the path $f$ "joins" $x$ and $y$. * A metric space $X$ is **path-connected** if any two points in $X$ can be joined by a path in $X$. * For example, the unit circle in $(\mathbb{R}^{2},d_{2})$ is connected; $\mathbb{R}\setminus \{ 0 \}$ is not connected, but $\mathbb{R}^{2}\setminus \{ 0 \}$ is. * Being connected by a path is an equivalent relation; the partition of the space into path-connected subsets is called **path-components**. * Path-connectedness implies connectedness. The converse is not necessarily true. > Proof: if for contradiction $X$ is path-connected but not connected, then there is continuous function $f: X \to \{ 0,1 \}$ not constant, say $\exists x,y \in X:f(x)\ne f(y)$. > But path-connectedness gives a path $\rho$ joining $x,y$, so $f \circ \rho$ gives a continuous, non-constant map $[0,1] \to \{ 0,1 \}$, a contradiction to $[0,1]$ being connected. > > Counterexample to the converse: google "Topologist's sine curve". * Nevertheless, connected normed vector spaces are path-connected. > Proof: assume for contradiction the normed vector space $V$ is connected but not path-connected. Then there are multiple path-components, and we shall show that they disconnect $X$. Since partition is immediate, it remains to show that these components are open. > > Take any $v \in V$ and its path-component $C_{v}$. Since $V$ itself is open, there is $\epsilon: B(v,\epsilon) \subseteq V$. To see that $B(v, \epsilon) \subseteq C_{v}$, fix any $u \in B(v,\epsilon)$ and $u,v$ can be joined with the path $p(\lambda)=\lambda u+(1-\lambda)v$. Hence the ball $B(a, \epsilon) \subseteq C_{a}$, the path-component is open. ## Sequential Compactness *Big idea: sequential compactness is the generalization of Bolzano-Weierstrass Theorem in $\mathbb{R}$ (that all bounded real sequences have a convergent subsequence).* * A metric space $X$ is **sequentially compact** if any sequence $(x_{n}) \subseteq X$ has a convergent subsequence. * For example, by Bolzano-Weierstrass Theorem in the reals, any closed, bounded subset of $\mathbb{R}$ is sequentially compact. ### Criteria for Sequential Compactness * A sequentially compact subspace $A \subseteq X$ must be bounded and closed; the converse is not true in general. > Proof: suppose $A$ is sequentially compact. > <u>Closedness</u>: if for contradiction $A$ is not closed, there is a limit point $a \not\in A$, and being a limit point, it has a sequence $A \supseteq (a_{n}) \to a$ in $X$. Then any subsequence $(a_{n_{k}}) \to a \not \in A$, a contradiction. > > <u>Boundedness</u>: If for contradiction $A$ is unbounded, then from an arbitrary point $p \in A$, for any distance $n \in \mathbb{N}$, there is a point further away: $\exists a_{n} \in A: d(p,a_{n})>n$. Then the sequence $(a_{n})$ has no convergent subsequence (any "limit" $a$ would have $d(a,a_{n})>d(a_{n},p) - d(p,a) \to \infty$). > > Counterexample to the converse: $(0,1)$ is closed and bounded *in itself*, but is not sequentially compact. * Sequential compactness is preserved by continuous maps. * In particular, <u>sequential compactness is shared by homeomorphic spaces</u>. > Proof: say $f: X \to Y$ continuous, and $X$ is sequentially compact. Take any $(y_{n}) \subseteq Y$, with preimage $(x_{n})\equiv(f^{-1}(y_{n}))$. Sequential compactness of $X$ gives subsequence $(x_{n_{k}}) \to x \in X$. > Continuity of $f$ then guarantees $y_{n_{k}}=f(x_{n_{k}}) \to f(x) \in Y$. Hence $(y_{n}) \subseteq Y$ has a convergent subsequence. * Products of sequentially compact spaces are also sequentially compact. > Proof: Take $X, Y$ sequentially compact and sequences $(x_{n})\subseteq X,(y_{n}) \subseteq Y$. > Then $X$ sequentially compact gives a convergent subsequence $(x_{n_{k}}) \to x$. Take $(\gamma_{k})=(y_{n_{k}}) \subseteq Y$, then sequential compactness of $Y$ gives $(\gamma_{k_{r}}) \to y$ convergent. > So $(x_{n_{k_{r}}},y_{n_{k_{r}}}) \to (x,y)$ is a convergent subsequence of $((x_{n}, y_{n}))$, so $X \times Y$ is sequentially compact. * **Bolzano-Weierstrass Theorem on** $\mathbb{R}^n$: any closed, bounded subset of $\mathbb{R}^n$ is sequentially compact. > Proof: say $X \subseteq \mathbb{R}^n$ closed and bounded in $[-M,M]^n$. Then by the previous result, $[-M,M]^n$ is sequentially compact. Since $X$ is a closed subset of it, $X$ is also sequentially compact. * **Heine-Cantor Theorem**: any continuous map from a sequentially compact space must be uniformly continuous. * in particular, any continuous map $f: X \to \mathbb{R}$, $X$ sequentially compact, must be uniformly continuous. > Proof: say there is a continuous map $f:X \to Y$, $X$ is sequentially compact. Suppose for contradiction $f$ is not uniformly continuous. > So there $\exists\epsilon>0$, for which any $\delta=1 / n, n \in \mathbb{N}$, there is exception $(a_{n},b_{n}):d_{X}(a_{n},b_{n})<\delta$, but $d_{Y}(f(a_{n}),f(b_{n})) \ge \epsilon$. > > By sequential compactness, there is a subsequence $(a_{n_{k}},b_{n_{k}}) \to (l,l)$, where the limits are both $l$ since $d_{X}(a_{n},b_{n}) \to 0$. But then continuity of $f$ requires $(f(a_{n_{k}}),f(b_{n_{k}})) \to (f(l),f(l))$, contradictory to $d_{Y}(f(a_{n}),f(b_{n})) \ge \epsilon$. ### Sequential Compactness and Completeness * Sequential compactness implies completeness and boundedness. > Proof: take $X$ sequentially compact and Cauchy sequence $(x_{k}) \subseteq X$ with convergent subsequence $(x_{k_{r}}) \to x$. Boundedness is already shown in the previous section. > > For completeness, we shall show that $(x_{n}) \to x$, the big idea being that since the sequence is Cauchy, the convergent subsequence wil "drag" the entire sequence with it towards the limit. > > Given $\epsilon >0$, take the indices where the subsequence grows close enough to the limit: $K=\{ k_{r}: d(x_{k_{r}},x) < \epsilon \}$. Cauchy-ness gives $l: \forall m,n \ge l, d(x_{m}, x_{n})< \epsilon$. In particular, take $m \in K$, so that $x_{m}$ can "drag" $x_{n}$ to the limit:$d(x,x_{n})<d(x,x_{m})+d(x_{n},x_{m}) < 2\epsilon$So $(x_{n}) \to x$, and the space is complete. - The converse is not true in general: in $\bar{B}(0,1) \subseteq (B(\mathbb{R}),d_{\infty})$ is bounded and complete, but not sequentially compact. > Proof: > <u>Boundedness</u> is immediate. > > It is <u>complete</u> is because of $\bar{B}(0,1)$ being a closed subset of $B(\mathbb{R})$, a complete metric space. > > It is <u>not sequentially compact</u> because the sequence indicator functions $f_{n}=\chi_{[n,n+1]}$ always has $d_{\infty}(f_{n},f_{m})=\delta_{mn}$, so there is no Cauchy (hence convergent) subsequence. * A space is **totally bounded** if for any radius $r>0$, the space can be covered by finitely many open balls of radius $r$: $\exists \text{ finite } \{ x_{i} \} \subseteq X:X \subseteq\bigcup_{i} B(x_{i}, r)$ * A space is sequentially compact $\iff$ it is complete and totally bounded. > Proof: ($\Rightarrow$ totally bounded) suppose for contradiction there is a radius $\epsilon>0$ such that the space cannot be covered by finitely many balls of radius $\epsilon$. > Then inductively taking an element $x_{n}$ from each ball gives a sequence $(x_{n})$ with no Cauchy subsequences (each element is at least $\epsilon$ from each other), hence no convergent subsequences, a contradiction. > > ($\Leftarrow$) Given a sequence $\sigma=(x_{n}) \subseteq X$, we construct a Cauchy subsequence explicitly, then completeness will guarantee its convergence. > For $n \in \mathbb{N}$, let $\mathcal{B}_{n}$ be the finite family of balls of radius $2^{-n}$ that covers the space. > Start with $n=0$, and at least one of the balls $B_{0} \in \mathcal{B}_{0}$ contains an infinite number of terms of $\sigma$, so take the subsequence $\sigma_{0}=\sigma \cap B_{0}$. > Now increment to $n=1$, and one of the balls $B_{1} \in \mathcal{B}_{1}$ contains infinitely many terms of $\sigma_{0}$, giving a new subsequence $\sigma_{1} =\sigma_{0} \cap B_{1}$. > Repeat for each $n \in \mathbb{N}$, and we get subsequences $(\sigma_{n})$ where $\sigma_{i+1} \subseteq \sigma_{i} \cap B_{i+1}$, and $\text{radius }(B_{i})=2^{-i}$. > Then we can construct a new subsequence $(\chi_{n}) \subseteq (x_{n})$ by taking the $n$th term of $\sigma_{n}$. Since $(\chi_{n \ge k}) \subseteq B_{k}$, they can only be at most $2^{k-1}$ apart, so this subsequence is Cauchy. Then completeness gives $(\chi_{n}) \to \chi$, a convergent subsequence of $(x_{n})$. ### Arzelà-Ascoli Theorem: Sequential Compactness in Function Spaces *Big idea: $(C(X),d_{\infty})$ is not sequentially compact, but some extra restrictions could make it so.* * The space of continuous functions over a sequentially compact spase $X$, $C(X)$ is not sequentially compact: in $C(\mathbb{R})$, * The supremum of the functions can tend to infinity: $(f_{n})=(n)$ has no convergent subsequence. * The functions can be increasingly steep, so they are all distant from each other under $d_{\infty}$: consider the piecewise linear functions $(f_{n}): f_{n}=\begin{cases}0 & \text{if }x \le \frac{1}{n+1} \\ n(n+1)(x-\frac{1}{n+1}) &\text{if } \frac{1}{n+1}<x \le \frac{1}{n}\\ 1 &\text{if }x> \frac{1}{n} \end{cases}$where $d_{\infty}(f_{m},f_{n})=1$ when $m \ne n$. * However, Arzelà-Ascoli states that as long as these two types of exceptions are removed, the space is sequentially compact. * A family of functions $\{ f_{n} \}$ is **uniformly bounded** if they share the same bound: $\exists M: \forall n,\, \sup|f_{n}| \le M$. This rules out the tending-to-infinity type of exceptions. * A family of continuous functions $\{ f_{n} \}$ is **equicontinuous** if given $\epsilon >0$, the same $\delta$ suffices for the $\epsilon-\delta$ definition of continuity for all $f_{n}$ at all $x,y \in X$: $\forall \epsilon>0,\, \exists \delta>0:\forall n,\, (d(x,y)<\delta) \Rightarrow |f_{n}(x)-f_{n}(y)| < \epsilon$This rules out the increasingly-steep type of exceptions. * **Arzelà-Ascoli**: over a sequentially compact space $X$, if $F \subseteq C(X)$ is uniformly bounded and equicontinuous, then any sequence in $F$ has a subsequence convergent in $C(X)$. Hence if $F$ is closed, it is also sequentially compact. > Proof is not examinable. ## Compactness * In a metric space $X$, a **cover** of $A \subseteq X$ is a family of subsets $\{ U_{i} \subseteq X \}$ where $A \subseteq \bigcup_{i}U_{i}$. * For example, $\{ B(k,2): k \in \mathbb{Z} \}$ is a cover of $\mathbb{R}$. * The set $\{ U_{i_{k}} \} \subseteq \{ U_{i} \}$ is a **subcover** if it is still a cover of $A$; if it has finitely many members, it is a **finite subcover**. * For example, $\{ B(2k,2) \}$ is a subcover of the previous example. * Note that finiteness referrs to the number of "patches", not the size: $\{ (-\infty, \infty) \}$ is a finite cover of $\mathbb{R}$. * A cover is an **open cover** if all its family members $U_{i}$ are open in $X$. * The previous examples are open covers since $B(k,2)$ is open in $\mathbb{R}$. * A subset $A$ is **compact** if every open cover has a finite subcover. * For example, $(0,1)$ is not compact in $\mathbb{R}$, since the open cover $\left\{ \left( \frac{1}{n}, 1 \right) : n \in \mathbb{Z}^+\right\}$ has no finite subcover. ### Criteria for Compactness - If $X$ is compact, and $f:X \to Y$ is continuous, then $f(X)$ is also compact. - In particular homeomorphic spaces share compactness. > Proof: let $\mathcal{C}$ be an open cover of $f(X)$. Then $f^{-1}(\mathcal{C})\equiv \{ f^{-1}(C)\,|\, C \in \mathcal{ C} \}$ is an open cover of $X$, since open sets $C \in \mathcal{C}$ have open preimages under $f$ open. Then compactness of $X$ gives a finite subcover $\mathcal{B} \subseteq f^{-1}(\mathcal{C})$, and so $f(\mathcal{B}) \subseteq \mathcal{C}$ is a finite subcover of $f(X)$. * Compact subsets are bounded. * Hence for example $\mathbb{R}$ is not compact. > Proof: take $A \subseteq X$, $A$ compact, but for contradiction assume that around $p$, any radius $n \in \mathbb{N}$ has $a_{n} \in A: d(p,a_{n}) >n$. > Then the set $\{ B(p,n): n \in \mathbb{N} \}$ is an open cover of $A$ that has no finite subcover, a contradiction. ### Compactness Equals Sequential Compactness * Compactness implies sequential compactness in metric spaces. > This a proof from Sutherland (2009). Take $(x_{n})$ from a compact space $X$. We will show that it has a convergent subsequence. > > Lemma: given sequence $(x_{n})$ with $x:\forall \epsilon>0,\, B(x,\epsilon)$ contains infinitely many terms of $(x_{n})$, then there is a subsequence $(x_{n_{k}}) \to x$. > > Proof: ![[CompactToSeqCompact.png]] > > We construct the subsequence inductively; start with $x_{n_{1}}\in B(x,1)$, which exists by assumption. > > Then we construct $x_{n_{k+1}}$ given $x_{n_{k}}$: since $(x_{n \ge n_{k}})$ has infinitely many terms within $B\left( x, \frac{1}{k+1} \right)$, we can take a $x_{n_{k+1}}$ from that. > > So $(x_{n_{k}}) \to x$, and it is a subsequence of $(x_{n})$. > > We will use its contrapositive: that if there are no convergent subsequences in $(x_{n})$, then $\forall x,\,\exists \epsilon>0: B(x,\epsilon)$ contains only finitely many of $(x_{n})$. > > Proof: assume for contradiction $(x_{n})$ has no convergent subsequence. Then by the lemma, all $x$ has $\epsilon: B_{x}=B(x,\epsilon)$ contains finitely many terms of $(x_{n})$. > The open cover $\{ B_{x}: x \in X \}$ has no finite subcover, since any finite subset contains only finitely many terms of $(x_{n})$, hence not covering it, a contradiction. * Sequential compactness implies compactness. > Proof is not examinable. ### Heine-Borel Theorem: Compactness in the Reals * **Heine Borel**: any closed, bounded interval $[a,b] \subset \mathbb{R}$ is compact. > Obviously $[a,b]$ is totally bounded and complete, hence sequentially compact, hence compact. Instead of this, Oxford wants the direct proof below. FML. > > Proof: given an open cover $\mathcal{U}=\{ U_{i} \}$, define the set $S= \big\{ c \in [a,b]: [a,c] \text{ has a finite subcover} \big\}$ and $s=\sup S$. We will show that $s=b$. ![[HeinBorelDiagram.png]] > Denote $U^{(x)} \in \mathcal{U}$ as some patch in the cover containing $x$. > Obviously $s> a$, since $\exists \epsilon>0: B(a,\epsilon) \subseteq U^{(a)}$, so $a+\frac{\epsilon}{2} \in S$. > Assume for contradiction that $s \ne b$. Then since $U^{(s)}$ open, so it contains $B(s, \eta)$ for some $|b-s|>\eta >0$. But then $s+ \frac{\eta}{2} \in S$, contradictory to the definition of $s=\sup S$. Hence $b = \sup S$. > There is $r:B(b,r) \subseteq U^{(b)}$. Since $b$ is $\sup S$, the approximation theorem gives $c \in S: |b-c|< r$. > Then $[a,c]$ has a finite subcover $\mathcal{U}_{[a,c]} \subseteq \mathcal{U}$. So the union $\mathcal{U}_{[a,c]} \cup \{ U^{(b)}\}$ is a finite subcover of $\mathcal{U}$.