2.6 A little topology  (Page 3/3)

We give next a result that clarifies to some extent the connection between open sets and closed sets. Always remember that there are sets that are neither open nor closed, and just because a set is not open does not mean that it is closed.

A subset $S$ of $C$ ( $R$ ) is open if and only if its complement $\tilde{S}=C\setminus S$ ( $R\setminus S$ ) is closed.

First, assume that $S$ is open, and let us show that $\tilde{S}$ is closed. Suppose not. We will derive a contradiction.Suppose then that there is a sequence $\left\{x_n\right\}$ of elements of $\tilde{S}$ that converges to a number $x$ that is not in $\tilde{S};$ i.e., $x$ is an element of $S.$ Since every element of $S$ is an interior point of $S,$ there must exist an $ϵ>0$ such that the entire disk $B_{ϵ}\left(x\right)$ (or interval $(x-ϵ,x+ϵ)$ ) is a subset of $S.$ Now, since $x=lim{x}_n,$ there must exist an $N$ such that $|x_n-x|<ϵ$ for every $n\ge N.$ In particular, $|x_N-x|<ϵ;$ i.e., $x_N$ belongs to $B_{ϵ}\left(x\right)$ (or $(x-ϵ,x+ϵ)$ ). This implies that $x_N\in S.$ But $x_N\in \tilde{S},$ and this is a contradiction. Hence, if $S$ is open, then $\tilde{S}$ is closed.

Conversely, assume that $\tilde{S}$ is closed, and let us show that $S$ must be open. Again we argue by contradiction.Thus, assuming that $S$ is not open, there must exist a point $x\in S$ that is not an interior point of $S.$ Hence, for every $ϵ>0$ the disk $B_{ϵ}\left(x\right)$ (or interval $(x-ϵ,x+ϵ)$ ) is not entirely contained in $S.$ So, for each positive integer $n,$ there must exist a point $x_n$ such that $|x_n-x|<1/n$ and $x_n\notin S.$ It follows then that $x=lim{x}_n,$ and that each $x_n\in \tilde{S}.$ Since $\tilde{S}$ is a closed set, we must have that $x\in \tilde{S}.$ But $x\in S,$ and we have arrived at the desired contradiction. Hence, if $\tilde{S}$ is closed, then $S$ is open, and the theorem is proved.

The theorem below, the famous Heine-Borel Theorem, gives an equivalent and different description of closed and bounded sets.This description is in terms of open sets, whereas the original definitions were interms of limit points.Any time we can find two very different descriptions of the same phenomenon, we have found something useful.

Let $S$ be a subset of $C$ (respectively $R).$ By an open cover of $S$ we mean a sequence $\left\{U_n\right\}$ of open subsets of $C$ (respectively $R$ ) such that $S\subseteq \cup U_n;$ i.e., for every $x\in S$ there exists an $n$ such that $x\in U_n.$

A subset $S$ of $C$ (respectively $R$ ) is called compact , or is said to satisfy the Heine-Borel property , if every open cover of $S$ has a finite subcover. That is, if $\left\{U_n\right\}$ is an open cover of $S,$ then there exists an integer $N$ such that $S\subseteq {\cup }_{n=1}^N{U}_n.$ In other words, only a finite number of the open sets are necessary to cover $S.$

REMARK The definition we have given here for a set being compact is a little less general from the one found in books on topology.We have restricted the notion of an open cover to be a sequence of open sets, while in the general setting an opencover is just a collection of open sets. The distinction between a sequence of open sets and a collection of open setsis genuine in general topology, but it can be disregarded in the case of the topological spaces $R$ and $C.$

Heine-borel theorem

A subset $S$ of $C$ (respectively $R$ ) is compact if and only if it is a closed and bounded set.

We prove this theorem for subsets $S$ of $C,$ and leave the proof for subsets of $R$ to the exercises.

Suppose first that $S\subseteq C$ is compact, i.e., satisfies the Heine-Borel property. For each positive integer $n,$ define $U_n$ to be the open set $B_n\left(0\right).$ Then $S\subseteq \cup U_n,$ because $C=\cup U_n.$ Hence, by the Heine-Borel property, there must exist an $N$ such that $S\subseteq {\cup }_{n=1}^N{U}_n.$ But then $S\subseteq B_N\left(0\right),$ implying that $S$ is bounded. Indeed, $\left|x\right|\le N$ for all $x\in S.$

Next, still assuming that $S$ is compact, we will show that $S$ is closed by showing that $\tilde{S}$ is open. Thus, let $x$ be an element of $\tilde{S}.$ For each positive integer $n,$ define $U_n$ to be the complement of the closed set $\overline{B_{1/n}\left(x\right)}.$ Then each $U_n$ is an open set by Theorem 2.12, and we claim that $\left\{U_n\right\}$ is an open cover of $S.$ Indeed, if $y\in S,$ then $y\ne x,$ and $|y-x|>0.$ Choose an $n$ so that $1/n<|y-x|.$ Then $y\notin \overline{B_{1/n}\left(x\right)},$ implying that $y\in U_n.$ This proves our claim that $\left\{U_n\right\}$ is an open cover of $S.$ Now, by the Heine-Borel property, there exists an $N$ such that $S\subseteq {\cup }_{n=1}^N{U}_n.$ But this implies that for every $z\in S$ we must have $|z-x|\ge 1/N,$ and this implies that the disk $B_{1/N}\left(x\right)$ is entirely contained in $\tilde{S}.$ Therefore, every element $x$ of $\tilde{S}$ is an interior point of $\tilde{S}.$ So, $\tilde{S}$ is open, whence $S$ is closed. This finishes the proof that compact sets are necessarily closed and bounded.

Conversely, assume that $S$ is both closed and bounded. We must show that $S$ satisfies the Heine-Borel property. Suppose not. Then, there exists an open cover $\left\{U_n\right\}$ that has no finite subcover. So, for each positive integer $n$ there must exist an element $x_n\in S$ for which $x_n\notin {\cup }_{k=1}^n{U}_k.$ Otherwise, there would be a finite subcover. By [link] , there exists a subsequence $\left\{x_{n_j}\right\}$ of $\left\{x_n\right\}$ that converges to an element $x$ of $S.$ Now, because $\left\{U_n\right\}$ is an open cover of $S,$ there must exist an $N$ such that $x\in U_N.$ Because $U_N$ is open, there exists an $ϵ>0$ so that the entire disk $B_{ϵ}\left(x\right)$ is contained in $U_N.$ Since $x=lim{x}_{n_j},$ there exists a $J$ so that $|x_{n_j}-x|<ϵ$ if $j\ge J.$ Therefore, if $j\ge J,$ then $x_{n_j}\in U_N.$ But the sequence $\left\{n_j\right\}$ is strictly increasing, so that there exists a $j^{\text{'}}\ge J$ such that $n_{j^{\text{'}}}>N,$ and by the choice of the point $x_{n_{j^{\text{'}}}},$ we know that $x_{n_{j^{\text{'}}}}\notin {\cup }_{k=1}^N{U}_k.$ We have arrived at a contradiction, and so the second half of the theorem is proved.