1.5 Topological concepts

Interiors and closures

Definition 1 Let $(X,d)$ be a metric space. The ball $B(x_0,ϵ)$ centered at $x_0\in X$ and of radius $ϵ\ge 0$ is defined as $B(x_0,ϵ)=\{x\in X:d(x,x_0)<ϵ\}$ .

If $(X,\parallel ·\parallel )$ is a metric space, then $B(x_0,ϵ)=\{x\in X:\parallel x-x_0\parallel <ϵ\}$ .

Example 1 In the metric space $({\mathbb{R}}^2,d_2)$ , the ball $B(p_0,ϵ)$ is a circle centered at $p_0$ and of radius $ϵ$ , as illustrated in [link] .

The following four definitions are fundamental in the study of topology.

Definition 2 Let $P\subset X$ where $(X,d)$ is a metric space. A vector $p_0\in P$ is an interior point of P if there exists $ϵ>0$ such that $B(p_0,ϵ)\subseteq P$ .

Intuitively, the notion of an interior point is a point that is not in the “boundary” of the set, as a ball around it is contained within the set.

Definition 3 The interior of a set $P$ is the collection of all the interior points of $P$ and is denoted as $P^{\circ }$ .

Intuitively, closure points are points that are arbitrarily close to the set $P$ ; note however that a closure point need not be in $P$ , but only have a sequence of elements of $P$ that converge to it.

Definition 4 A point $p_1\in X$ is a closure point of P if for all $ϵ>0$ we have that $B(p_1,ϵ)\cap P\ne \varnothing$ .

Definition 5 The closure of a set P is the set of all closure points of P denoted as $\overline{P}$ .

Open and closed sets

Topology is the study of open and closed sets, defined below.

Definition 6 A set P is said to be open if $P=P^{\circ }$ , i.e., every point in $P$ is an interior point of $P$ .

Definition 7 A set P is said to be closed if $P=\overline{P}$ , i.e., $P$ contains all its closure points.

Fact 1 Since all interior points of $P$ are in $P$ and every point in $P$ is a closure point of $P$ , we have that $P^{\circ }\subseteq P\subseteq \overline{P}$ .

Example 2 The following are examples of open and closed sets.

• The set $[a,b]=\{x\in \mathbb{R}:a\le x\le b\}$ is closed.
• The set $(a,b)=\{x\in \mathbb{R}:a<x<b\}$ is open. To show this, we must show that every point $x\in (a,b)$ is an interior point. Pick an arbitrary $x\in (a,b)$ , and define $ϵ=min\left(\frac{x-a}{2},,,\frac{b-x}{2}\right)$ . Then the ball $B(x,ϵ)=\left\{u,\in ,\mathbb{R},:,|,u,-,x,|,<,ϵ\right\}$ can be rewritten as the set of all points $u$ such that $-ϵ+x<u<ϵ+x$ . Using the definition of $ϵ$ , we have that if $u\in B(x,ϵ)$ then
  $$-\frac{x-a}{2}+x\le u\le \frac{b-x}{2}+x,$$
  or equivalently,
  $$\frac{x+a}{2}\le u\le \frac{b+x}{2}.$$
  Since $a<x<b$ , we have
  $$a<\frac{x+a}{2}\le u\le \frac{b+x}{2}<b,$$
  and so $u\in (a,b)$ . Since $u\in B(x,ϵ)$ was arbitrary, we then have $B(x,ϵ)\subseteq (a,b)$ and $x$ is an interior point of $(a,b)$ . Now since $x\in (a,b)$ was arbitrary, then the set $(a,b)$ is open.

Properties of open and closed sets

Theorem 1 ( $i$ ) If $A$ is open then $A^C$ is closed. ( $ii$ ) If $A$ is closed then $A^C$ is open.

Proof: ( $i$ ) We will prove by contradiction: Assume $A$ is open and $A^C$ is not closed, that is, there exists a closure point $x$ of $A^C$ such that $x\notin A^C$ , that is, $x\in A$ . Since $x$ is a closure point of $A^C$ , we have that for every $ϵ>0$ ,

Since $A$ is open and $x\in A$ , then $x$ is an interior point of $A$ , which means that there exists ${ϵ}_0>0$ such that $B(x,{ϵ}_0)\subseteq A$ , which means that $B(x,{ϵ}_0)\cap A^C=\varnothing$ , a contradiction with [link] . Therefore, we must have that $A^C$ is closed.

( $ii$ ) Assume $A$ is closed, which means that $A$ contains all its interior points. That means that if $x\in A^C$ then $x$ is not a closure point of $A$ , meaning that there for some ${ϵ}_0>0$ we must have $B(x,{ϵ}_0)\cap A=\varnothing$ . This means that $B(x,{ϵ}_0)\subseteq A^C$ , and so $x$ is an interior point of $A^C$ . Since $x$ was an arbitrary point in $A^C$ , this means that $A^C$ is open.

Proposition 1 The intersection of a finite number of open sets is open, and the union of an arbitrary collection of open sets is open.

Proof: We will limit the proof to two sets, which can be extended in each case using a proof by induction argument.

We first show that if $A_1$ , $A_2$ are open then $A_1\cap A_2$ is open, i.e., ${(A_1\cap A_2)}^O=A_1\cap A_2$ . Assume $x\in A_1\cap A_2$ ; then $x\in A_1$ and $x\in A_2$ . Since $A_1$ , $A_2$ are open then there exists ${ϵ}_1$ , ${ϵ}_2>0$ such that $B(x,{ϵ}_1)\subseteq A_1$ and $B(x,{ϵ}_2)\subseteq A_2$ . Set $ϵ=min({ϵ}_1,{ϵ}_2)$ ; then, $B(x,ϵ)\subseteq B(x,{ϵ}_1)$ and $B(x,ϵ)\subseteq B(x,{ϵ}_2)$ . By transitivity of inclusion, we have that $B(x,ϵ)\subseteq A_1$ and $B(x,ϵ)\subseteq A_2$ . Theferore, $B(x,ϵ)\subseteq A_1\cap A_2$ .

Next, we show that if $A_1$ , $A_2$ are open then $A_1\cup A_2$ is open, i.e., ${(A_1\cup A_2)}^O=A_1\cup A_2$ . Assume $x\in A_1\cup A_2$ ; then $x\in A_1$ or $A_2$ . If $x\in A$ , then there exists ${ϵ}_1>0$ s.t. $B(x,{ϵ}_1)\subseteq A_1\subseteq A_1\cup A_2$ . Similarly, if $x\in A_2$ , there exists ${ϵ}_2>0$ s.t. $B(x,{ϵ}_2)\subseteq A_2\subseteq A_1\cup A_2$ . So $x\in A_1\cup A_2$ is an interior point of $A_1\cup A_2$ and therefore ${(A_1\cup A_2)}^O=A_1\cup A_2$ .

Proposition 2 The union of a finite number of closed sets is closed. The intersection of an arbitrary collection of closed sets is closed.

The following useful properties are proven in “Optimization by Vector Space Methods” by David Luenberger, pages 25 and 38.

Proposition 3 If C is convex then its interior $C^O$ and closure $\overline{C}$ are convex.

Proposition 4 A subset of a Banach space is complete if and only if it is closed.

Proposition 5 Any finite dimensional subspace of a normed linear space is complete.

Why are Banach spaces useful? In optimization, we want to show that if an increasingly better solution can be found then an optimum must exist.