Lecture 4. Set Topology on Metric Spaces

While the real and complex numbers are the basis of analysis, they are very limited in at least two important ways. First they are one dimensional (as vector spaces over themselves), while many interesting and natural mathematical objects such as functions live in infinite dimensional spaces. Second they are linear (read carry a vector space structure), while again most objects are nonlinear, starting with the planet we live on. As approximation procedures are ubiquitous and essential to the solution of many important problems which live in either infinite dimensional or nonlinear (or both) spaces, we need a good replacement for the concept of absolute value (or modulus, in the complex case). The main purpose is of course the ability to consider convergence. This can be done in essentially three ways. From the more special to the more general, they are:
(a) Norms. These are length-measuring functions on vector spaces.
(b) Distances. These are distance measuring functions and can be defined on general sets.
(c) Topologies. These consist of a family of subsets of a given set, which encode the vague notion of proximity.

A norm on a vector space $X$ is a map $\|\cdot\|:X\to \mathbb{R}$ which satisfies the following additional properties:
(i) (Postivity) For any $x\in X$ it holds that $\| x\|\geq 0$ with equality only for $x=0$.
(ii) (positive Homogeneity) For any scalar $\lambda$ and any vector $x\in X$, it holds that $\| \lambda x\|=|\lambda|\| x\|$.
(iii) (Triangle Inequality) For any two vectors $x,y\in X$ it holds that $\| x+y\|\leq \| x\|+\| y\|$.

These properties clearly are all satisfied by the absolute value on $\mathbb{R}$ and they are motivated by it. They also make intuitive sense as we "expect'' non-zero vectors to have some (non-negative) length, stretched/shrunk vectors to have accordingly stretched/shrunk lengths, and taking detours along different directions to make our journey longer. While very useful in the context of finite and infinite dimensional vector spaces, they cannot possibly provide any help in measuring distances between points on, e.g. a sphere. We observe, however, that given two points in a normed vector space $(X,\|\cdot\|)$ we can think of $\|x-y\|$ as the distance between them. We thus obtain a map $$ d:X\times X\to[0,\infty),\: (x,y)\to\| x-y\| $$ which satisfies the properties of a distance function as described in the beginning of Lecture 3.
Once it is realized that these properties encode what we would like a distance-measuring function to satisfy, it is only a small leap to give up the vector space structure and allow the function $d$ to be defined on general sets. In the context of vector spaces, we can check convergence of a sequence $(x_n)_{n\in\mathbb{N}}$ towards a limit $x_\infty$ by verifying whether the length of $x_n-x_\infty$ converges to zero. In the language of distances, this amounts to verifying whether the distance between $x_n$ and $x_\infty$ tends to zero. A natural way to visualize this is to consider the set $\mathbb{B}(x_\infty,\varepsilon)$ of all points which lie at most a certain given distance $\varepsilon>0$ from $x_\infty$, i.e. $$ \mathbb{B}(x_\infty,\varepsilon)=\{ x\in X\, :\, d(x,x_\infty)\leq\varepsilon\} $$ and (equivalently) rephrasing convergence by requiring that all but possibly finitely many elements of the sequence to lie in $\mathbb{B}(x_\infty,\varepsilon)$ no matter how small $\varepsilon>0$ gets. This point of views yields the last generalization, where one only requires existence of a family of sets (called open sets) which allow one to zoom in on any point of the space in a similar way as the balls above do. While we won't dwell on this, we remark that basic properties necessary for the family of "open'' sets $\mathcal{O}\subset 2^X$ (also called topology) are
(i) $\emptyset,X\in \mathcal{O}$.
(ii) $\bigcup_{\alpha\in A}O_\alpha\in \mathcal{O}$ whenever $O_\alpha\in\mathcal{O}$ for each $\alpha\in A$ and arbitrary index set $A$.
(iii) $\bigcap _{k=1}^nO_k\in \mathcal{O}$ for any finite number $n\in \mathbb{N}$ of sets $O_k\in\mathcal{O}$.

Convergence to a limit $x_\infty$ would in this context be defined by the requirement that all but finitely many element of the sequence belong to any given open set containing $x_\infty$. You might want to ponder the reasons why these are the conditions which lead to an acceptable notion of convergence. Notice that a norm always gives rise to a distance function, which, in turn, always yields a family of open sets as explained later in this lecture. It is also worthwhile mentioning that distance functions are very useful when dealing with nonlinear objects (smooth manifolds in particular), whereas norms and topologies are when dealing with vector spaces. Observe also that there are useful notions of convergence on infinite dimensional spaces which can only be defined via topologies (such that there is no norm or distance by which to define its open sets). When solving equations involving unknown functions (such as in the case of differential equations) one is often lead to sequence of functions $(f)_{n\in\mathbb{N}}$ on the convergence of which the existence of solutions often depends. Notice that when the functions $f:M\to X$ map from a manifold (think a sphere) $M$ into a vector space $X$, functions can be added and stretched by adding and stretching their values (thus obtainining a vector space structure amenable to the introduction of a norm). On the other hand when the image space is a (nonlinear) manifold, then addition and scalar multiplication are no longer available and the space of functions itself is no longer linear (thus making it impossible to measure distance between functions with a norm).

Basic Definitions

Definitions

Let $(X,d)$ be a metric space and $E$ a subset of $X$. Then
(i) $x \in E$ is called interior point of $E$ if there is $r>0$ such that $\mathbb{B}(x,r)\subset E$. The interior $\overset{\circ}{E}$ of $E$ is then defined as $$ \overset{\circ}{E}=\hbox{int}(E) = \{x \in E\, |\, x \text{ is an interior point of } E\}. $$
(ii) $E$ is said to be open iff every point of $E$ is an interior point, i.e. iff $\hbox{int}(E)= E$.
(iii) $x\in E$ is called isolated point if there is $r>0$ s.t. $\mathbb{B}(x,r) \cap E = \{x\}$.
(iv) $x\in X$ is called limit point of $E$ iff, for any $\epsilon>0$, there is $y \in \mathbb{B}(x,\epsilon) \cap E,\: y \ne x$. We denote the set of all limit points of $E$ by $E'$.
(v) $E$ is said to be closed whenever $E' \subset E$. The closure $\overline{E}$ of $E$ is given by $E' \cup E$. Notice that $E \setminus E' $ contains all isolated points of $E$.
(vi) $E$ is said to be a perfect set if $E = E'$.
(vii) The exterior of $E$ is given by ext$(E)$=int$(E^c)$.
(viii) The boundary of $E$ is defined by $\partial E=\overline{E}\setminus\text{int}(E)$.
(ix) $E$ is a compact set iff every open cover for $E$ contains a finite subcover. More precisely iff \begin{equation*} E\subset\cup _{\lambda\in\Lambda}O_\lambda\text{ for open sets }O_\lambda\text{ implies that }\exists\: \lambda_1,\dots,\lambda_n\in\Lambda\text{ such that } E\subset\cup _{k=1}^n O_{\lambda _k}\, . \end{equation*} (x) Two sets $A,B \subset X$ are called separated iff $A\cap \overline{B} = \overline{A} \cap B = \emptyset$.
(xi) A set $E\subset X$ is called connected if there are no non-empty separated sets $A$ and $B$ such that $E = A \cup B$.
(xii) A set $E\subset X$ is said to be dense in $X$ iff $\overline{E}=X$.

Exercise

Proposition

Let $(X,d)$ be a metric space. Then
(i) The ball $\mathbb{B}(x,r)$ is an open set.
(ii) $E$ is open if and only if $E^c$ is closed.
(iii) Any union of open sets is open.
(iv) Any intersection of closed sets is closed.

Example

It suffices to prove that $y$ is an interior point of $\mathbb{B}(x,r)$ for any $y \in\mathbb{B}(x,r)$. We claim that $\mathbb{B}(y,\epsilon) \subset \mathbb{B}(x,r)$ for $\epsilon = r- d(x,y)>0$. Indeed \begin{equation*} d(z,x) \le d(z,y) + d(x,y) < \epsilon + d(x,y)= r- d(x,y) + d(x,y) = r,\: z\in\mathbb{B}(y,\epsilon) \end{equation*}

Exercise

Remark

Example

To prove $\hbox{int}(E)$ is open, it suffices to prove that $x$ is an interior point of $\hbox{int}(E)$ for any $ x \in \hbox{int}(E)$. Let $x_0\in \hbox{int}(E)$. By definition there is $r>0$ such that $\mathbb{B}(x_0, r)\subset E$. It follows that $$ \mathbb{B}(x_0, r)=\hbox{int}(\mathbb{B}(x_0, r))\subset \hbox{int}(E)\, , $$ and $x_0$ is an interior point of $\hbox{int}(E)$ as desired.

Example

(i) We show that $E' = [0,1]$. Let $x \in (0,1]$ and $\epsilon>0$. Then by density of $\mathbb{Q} $ in $\mathbb{R}$, there exists a $r$ such that $$ r \in (x-\epsilon, x)\cap(x/2, x) \cap [0,1]\cap \mathbb{Q} \subset E\, . $$ By definition, $x\in E'$. Also $0 = \lim_{n \to \infty} 1/n$, where $1/n \in E$. It follows that $0 \in E'$ and we are done.
(ii) We show that $\hbox{int}(E) = \emptyset$. Take $x\in E$, choose $n \in\mathbb{N}$ with $n \ge 2(r+1)$, and let $y = x + \sqrt{2}/n$. Then $$ (y-x) =\sqrt{2}/n < r\text{ and }y \not \in E\, . $$ Therefore, $x \not\in \hbox{int}(E)$ and $\hbox{int}(E)= \emptyset$.

Compactness

Definition (Boundedness)

Theorem

(i) A compact set must be closed and bounded.
(ii) A closed subset of a compact set is compact.

(i) Let $E$ be a compact set. If $E^c=\emptyset$, then the claim follows. If not, let $x_0\in E^c$. Now, if $x\in E$, then $d(x,x_0)>0$ and there is $n\in \mathbb{N}$ such that $d(x,x_0)>1/n$, i.e., such that $x\in \overline{\mathbb{B}}(x_0,1/n)^c$. It follows that $$ E\subset\bigcup _{n\in \mathbb{N}}\overline{\mathbb{B}}(x_0,1/n)^c, $$ and the cover must possess a finite subcover. Therefore there is $n_0\in \mathbb{N}$ such that $E\subset \overline{\mathbb{B}}(x_0,1/n_0)^c$ or, equivalently, such that $$ \mathbb{B}(x_0,1/n_0)\subset E^c\, . $$ This shows that $E^c$ is open and thus $E$ is closed. Given $x_0\in E$, it always holds that $$ E\subset \bigcup_{n\in \mathbb{N}}\mathbb{B}(x_0,n). $$ Since this is an open cover, a finite subcover suffices and therefore $E\subset \mathbb{B}(x_0,M)$ for some $M\in \mathbb{N}$, which yields the desired boundedness.
(ii) Let $E$ be compact and $F$ be a closed subset of $E$. Then any open cover $\mathcal{O}=\{O_\lambda: \lambda\in \Lambda\}$ of $F$ yields the open cover $$ \{O_\lambda: \lambda\in \Lambda\}\cup \{X\setminus F\} $$ of $E$. Latter admits a finite subcover. Removing $X\setminus F$ from it if necessary, yields a finite subcover of $\mathcal{O}$ for $F$.

Example

If $\cap_{n=1}^\infty E_n = \emptyset$, then $$ X = \left( \cap_{n=1}^\infty E_n \right) ^c = \cup_{n=1}^\infty E_n^c. $$ The sets $E_n$ are closed since they are compact and thus their complements $E_n^c$ are open. Since $E_1 \subset X \subset \cup_{n\in \mathbb{N} }E_n^c$, it follows that $\{E_n^c: n=1,2,\dots\}$ is an open cover for $E_1$. By compactness it must possess a finite subcover $\{E_{n_j}^c: j=1,...,k\}$ with $n_1\le n_2\le \cdots\le n_k$ and thus $$ E_{n_k}\subset E_1\subset \cup_{j=1}^k E_{n_j}^c\subset E_{n_k}^c\, , $$ a contradiction. Therefore $\cap_{n=1}^\infty E_n \ne \emptyset$.

Exercises

(a) Try to give a proof on your own. You can also find one in Lecture 6 just prior to the Arzéla-Ascoli Theorem.
(b) Let $(x_n)_{n\in\mathbb{N}}$ be a non convergent Cauchy sequence in the set $K$. We will show that $K$ has an open cover with no finite subcover and, thus, can not be compact. As the sequence does not converge, given any $x\in K$, an $\varepsilon(x)>0$ can be found such that $$ \forall\: N\in \mathbb{N} \text{ there is }n_N\geq N\text{ with }d(x_{n_N},x)> 2\varepsilon(x). $$ As the sequence is Cauchy, there is $M(x)\in \mathbb{N}$ such that $$ d(x_n,x_m)\leq \varepsilon(x)\text{ whenever }m,n\geq M(x), $$ which then entails that $$ d(x_n,x)\geq d(x_{n_{M(x)}},x)-d(x_{n_{M(x)}},x_n)\geq \varepsilon(x)\text{ for }n\geq M(x). $$ It follows that $ \mathbb{B}\bigl(x, \varepsilon(x)\bigr)$ contains at most finitely many elements of the sequence $(x_n)_{n\in\mathbb{N}}$. As a consequence any finite collection of such balls will also only contain finitely many elements of the sequences. It can be concluded that the cover $$ \big\{ \mathbb{B}\bigl(x, \varepsilon(x)\bigr)\, \big |\, x\in K\big\} $$ of $K$ contains no finite subcover. Indeed such a finite subcover would not even cover all the elements of the sequence $(x_n)_{n\in\mathbb{N}}$.
(c) Let $K$ be compact and let $(x_n)_{n\in\mathbb{N}}$ be a sequence in $K$. Given $m\in\mathbb{N}$, $K$ can be covered by the open balls $\mathbb{B}(y, 1/2m)$, where $y$ runs through $K$. This open cover necessarily has a finite subcover so, that $y^m_1,\dots,y^m_{n_m}\in K$ can be founds such that $$ K\subset \bigcup_{k=1,\dots,n_m}\mathbb{B}(y^m_k,1/2m) $$ It follows that at least one of these balls contains infinitely many elements of the sequence $(x_n)_{n\in\mathbb{N}}$. Starting with $m=1$, let $(x_{n^1_j})_{j\in \mathbb{N}}$ be an infinite subsequence of elements in one of the balls of the size $1/2$. For $m=2$, there is subsequence $(x_{n^2_j})_{j\in \mathbb{N}}$ of $(x_{n^1_j})_{j\in \mathbb{N}}$ contained in one of the balls of size $1/4$. Continuing in this way, it is possible to construct a subsequence $(x_{n^m_j})_{j\in \mathbb{N}}$ contained in one of the balls of size $2^{-m-1}$. Taking the diagonal sequence $(y_j=x_{n^j_j})_{j\in \mathbb{N}}$, we obtain a subsequence of the original sequence, which has the Cauchy property. Indeed, by construction, we have that $$ d(y_j,y_k)\leq \max\{ 2^{-j-1},2^{-k-1}\},\: j,k\in \mathbb{N}. $$ It therefore has a limit as desired since $K$ is complete by (b).
Now assume that sequences in $K$ always possess convergent subsequences and give yourself an arbitrary open cover $\{ O_\alpha \,|\, \alpha \in A\}$ of $K$. A finite subcover needs to be found. We claim that there exists $ \varepsilon>0$ such that, for any given $x\in K$, there exists $\alpha(x)$ such that $\mathbb{B}(x, \varepsilon)\subset O_{\alpha(x)}$. Indeed, if that were not the case, then, given any $n\in \mathbb{N}$, $x_n$ could be found such $$ \mathbb{B}(x_n,1/n)\not\subset O_\alpha \text{ for }\alpha\in A. $$ By assumption, the sequence $(x_n)_{n\in\mathbb{N}}$ would have a convergent subsequence with limit $x_\infty\in K$, which, for simplicity, we relabel by the same name. Now, $x_\infty\in O_{\alpha_\infty}$ for some $\alpha _\infty\in A$ and $ \delta>0$ exists with $$ \mathbb{B}(x_\infty,\delta)\subset O_{\alpha_\infty}. $$ On the other hand convergence of the sequence yields $N\in \mathbb{N}$ such that $$ d(x_n,x_\infty)<\delta /2\text{ for }n\geq N. $$ A simple argument based on the triangle inequality then shows that $$ \mathbb{B}(x_n,1/n)\subset O_{\alpha_\infty}, $$ for $n$ large enough, contradicting the very construction of the balls. Next observe that, for any given $ \varepsilon>0$, any set $S\subset K$ with $$ x,y\in S\Longrightarrow d(x,y)\geq \varepsilon, $$ must be finite. If that were not the case, it would be straightforward to obtain a sequence with no convergent subsequence. We can now conclude the proof. We know $\varepsilon>0$ exists such that every ball centered in $K$ is contained in one of the open sets of the cover. Pick any $x_0\in K$. Now, either $K\subset \mathbb{B}(x_0,\varepsilon)$ or there is $x_1\notin \mathbb{B}(x_0,\varepsilon)$. Again, either $K\subset \cup _{j=0,1}\mathbb{B}(x_j,\varepsilon)$ or there is $x_2\notin \cup _{j=0,1}\mathbb{B}(x_j,\varepsilon)$. This procedure must stop after finitely many steps, say $M$, since it would otherwise lead to a set $S$ of the above type with infinitely many elements. In conclusion, we have that $$ K\subset \bigcup_{j=1,\dots,M}\mathbb{B}(x_j,\varepsilon)\subset \bigcup_{j=1,\dots,M}O_{\alpha_j}, $$ where $ \alpha _j$ is chosen such that $ \mathbb{B}(x_j,\varepsilon) \subset O_{\alpha_j}$, which gives the desired finite subcover.

Finite sets are arguably the simplest sets to understand. Any sequence in a finite set, will have a convergent subsequence! Indeed, if $(x_n)_{n\in \mathbb{N}}$ is a sequence in a finite set $S=\{ s_1,\dots, s_n\}$, then at least one of the elements has to repeat infinitely many times, thus yielding the desired convergent subsequence. You proved above that compact sets share this useful property: any sequence in a compact set has a convergent subsequence. Establishing compactness is one of the most, if not the most, widespread method to obtain convergence of a subsequence: often one starts with a problem that is not directly solvable and replaces it with a sequence of simpler problems that "converge" to it. The solution of the approximating problems yields a sequence. If a subsequence converges, then one has a candidate solution for the original problem. In many cases the unknown lives in an infinite dimensional space, hence understanding compactness in infinite dimensional spaces is very important. One example is considered in Lecture 6, where Arzéla-Ascoli's characterization of compactness is presented.

Compact Sets in $\mathbb{R}^n$

Theorem (Intervals)

Any $n$-dimensional interval of the form $$ I= [a_1, b_1] \times....\times [a_n,b_n] $$ is a compact subset of $\mathbb{R}^n$ .

We give a proof for the case $n=1$. The proof is similar when $n>1$. Let $\{ O_\lambda : \lambda \in \Lambda\}$ be an open cover of $I_0=[a_0,b_0]$ and suppose, towards a contradiction, that there is no finite subcover for $I_0$. Then, at least one of $$ [a_0, \frac{a_0+b_0}{2}]\text{ or }[\frac{a_0+b_0}{2}, b_0] $$ has no finite subcover. Denote it by $I_1=[a_1, b_1]$. Then, $$ a_0\le a_1\le b_1\le b_0; \quad b_1-a_1=\frac{b_0-a_0}{2} $$ Repeating this splitting procedure recursively, we obtain a sequence of intervals $(I_k)_{k\in\mathbb{N}}$ with \begin{equation*} I_k=[a_k,b_k]\, ,\: a_{k-1}\le a_k\le b_k\le b_{k-1}\text{ and with }b_k-a_k=\frac{b_{k-1}-a_{k-1}}{2}=\frac{b-a}{2^k} \, ,\: k\geq 1\, . \end{equation*} It is clear that $(a_k)_{k\in\mathbb{N}}$ and $(b_k)_{k\in\mathbb{N}}$ are convergent sequences and that they converge to the same point $x_0\in I=[a, b]$. Necessarily $x_0\in O_{\lambda}$ for some $\lambda\in \Lambda$, hence there is $\epsilon>0$ with $$ (x_0-\epsilon, x_0+\epsilon)\subset O_{\lambda}\, . $$ As soon as $(b-a)2^{-k}<\epsilon$, we have that $I^k \subset (x_0-\epsilon, x_0+\epsilon)\subset O_\lambda$, a contradiciton to the fact that $I^k$ can not be covered by finitely many of the sets in $\{O_{\lambda}: \lambda\in \Lambda\}$. Therefore $I=[a_0, b_0]$ must be compact.

Corollary (Nestled Intervals)

Let a sequence $(I_k)_{k\in\mathbb{N}}$ of $n$-dimensional compact intervals satisfy $$ I_1\supset I_2\supset\dots\supset I_k \supset\dots\text{ and }\hbox{diam}(I_k) \to 0\text{ as }k \to\infty\, . $$ Then $\cap_{k=1}^\infty I_k = \{x_0\}$ for some $x_0\in \mathbb{R} ^n$.

Remark

Examples

Theorem (Heine-Borel)

Let $E \subset \mathbb{R}^n$. Then $$ E\text{ is compact }\iff E\text{ is closed and bounded.} $$

"$\Rightarrow$'' If $E$ is compact, then it is closed. Since the cover $$ E\subset\cup _{M\in \mathbb{N}}(-M,M)^n\, , $$ must have a finite subcover, we find $N\in \mathbb{N}$ such that $E\subset(-N,N)^n$. Thus $E$ is bounded as well.
"$\Leftarrow$'' If $E$ is bounded then we find $M>0$ such that $$ E\subset [-M,M]^n\, . $$ We therefore see that $E$ is a closed subset of the compact set $[-M,M]^n$ and, as such, itself a compact set.

Theorem (Weierstrass)

Every infinite bounded $E\subset\mathbb{R}^n$ has a limit point in $\mathbb{R}^n$.

Since $E$ is bounded, there is $M>0$ such that $E\subset I=[-M, M]^n$. Using the idea of the proof of Intervals' Theorem, one can obtain intervals $I^k$ containing infinitely many elements of $E$ and satisfying \begin{equation*} I_0=[-M,M]\, ,\: I_k \subset I_{k-1}\text{ for }k\in \mathbb{N}\text{ and }\hbox{diam}(I^k)\le \frac{2M\sqrt{n}}{2^k}\, . \end{equation*} Then $\cap _{k\in \mathbb{N}}I_k=\{x_0\}$ for some $x_0$ which is a limit point of $E$.

Connected and Convex Sets

Definition (Connected Sets)

$E$ is said to be connected if $E \ne A \cup B$ whenever $$ A \ne \emptyset\, ,\: B \ne \emptyset\, ,\: \bar{A}\cap B =\emptyset\, ,\: A\cap \bar{B}= \emptyset\, . $$

Theorem

Any non-empty connected set $E\subset\mathbb{R}$ is an interval, i.e. there are $x_1, x_2 \in \mathbb{R}$ with $x_1 < x_2$ such that $$ (x_1, x_2)\subset E\subset [x_1,x_2]\, . $$

Suppose that the conclusion is not true. Then there are $x,y \in E$ with $x < y$ and $z\ne E$ such that $x < z< y$. Let $$ A = (-\infty, z) \cap E\text{ and }B = (z, \infty) \cap E $$ so that $E=A\cup B$. We have that $$ \bar{A} \subset (-\infty, z] \cap \bar{E}\text{ and }\bar {B} \subset [z, \infty)\cap \bar{E}\, . $$ Thus $\bar{A} \cap B = \emptyset$ and $A \cap\bar B = \emptyset$. Since $x \in A$, $y\in B$, $A,B$ are not empty, separated, sets in $\mathbb{R}$ and $E= A\cup B$. It follows that $E$ is not connected and the theorem is proved.

Definitions

(i) For any $x,y \in \mathbb{R}^n$, we denote the line segment connecting $x$ to $y$ by $[x,y]$. Thus $$ [x,y]=\big\{ tx +(1-t) y\, |\, t\in[0,1]\big\}. $$ (ii) A set $E \subset \mathbb{R}^n$ is called convex if $[x,y] \subset E$ for all $x,y \in E$.

Examples