2.6 A little topology  (Page 2/3)

Let $S$ be a bounded and closed subset of $C$ . Then every sequence $\left\{x_n\right\}$ of elements of $S$ has a subsequence that converges to an element of $S.$

Let $\left\{x_n\right\}$ be a sequence in $S.$ Since $S$ is bounded, we know by [link] that there exists a subsequence $\left\{x_{n_k}\right\}$ of $\left\{x_n\right\}$ that converges to somenumber $x.$ Since each $x_{n_k}$ belongs to $S,$ it follows that $x$ is a limit point of $S.$ Finally, because $S$ is a closed subset of $C,$ it then follows that $x\in S.$

We have defined the concept of a closed set. Now let's give the definition of an open set.

Let $S$ be a subset of $C.$ A point $x\in S$ is called an interior point of $S$ if there exists an $ϵ>0$ such that the open disk $B_{ϵ}\left(x\right)$ of radius $ϵ$ around $x$ is entirely contained in $S.$ The set of all interior points of $S$ is denoted by $S^0$ and we call $S^0$ the interior of $S.$

A subset $S$ of $C$ is called an open subset of $C$ if every point of $S$ is an interior point of $S;$ i.e., if $S=S^0.$

Analogously, let $S$ be a subset of $R.$ A point $x\in S$ is called an interior point of $S$ if there exists an $ϵ>0$ such that the open interval $(x-ϵ,x+ϵ)$ is entirely contained in $S.$ Again, we denote the set of all interior points of $S$ by $S^0$ and call $S^0$ the interior of $S.$

A subset $S$ of $R$ is called an open subset of $R$ if every point of $S$ is an interior point of $S;$ i.e., if $S=S^0.$

We give next a useful application of the Bolzano-Weierstrass Theorem, or more precisely an application of [link] . This also provides some insight into the structure of open sets.

Let $S$ be a closed and bounded subset of $C,$ and suppose $S$ is a subset of an open set $U.$ Then there exists an $r>0$ such that the neighborhood $N_r\left(S\right)$ is contained in $U.$ That is, every open set containing a closed and bounded set $S$ actually contains a neighborhood of $S.$

If $S$ is just a singleton $\left\{x\right\},$ then this theorem is asserting nothing more than the fact that $x$ is in the interior of $U,$ which it is if $U$ is an open set. However, when $S$ is an infinite set, then the result is more subtle. We argue by contradiction. Thus, suppose there is no such $r>0$ for which $N_r\left(S\right)\subseteq U.$ then for each positive integer $n$ there must be a point $x_n$ that is not in $U,$ and a corresponding point $y_n\in S,$ such that $|x_n-y_n|<1/n.$ Otherwise, the number $r=1/n$ would satisfy the claim of the theorem. Now, because the $y_n$ 's all belong to $S,$ we know from [link] that a subsequence $\left\{y_{n_k}\right\}$ of the sequence $\left\{y_n\right\}$ must converge to a number $y\in S.$ Next, we see that

and this quantity tends to 0. Hence, the subsequence $\left\{x_{n_k}\right\}$ of the sequence $\left\{x_n\right\}$ also converges to $y.$

Finally, because $y$ belongs to $S$ and hence to the open set $U,$ we know that there must exist an $ϵ>0$ such that the entire disk $B_{ϵ}\left(y\right)\subseteq U.$ Then, since the subsequence $\left\{x_{n_k}\right\}$ converges to $y,$ there must exist a $n_k$ such that $|x_{n_k}-y|<ϵ,$ implying that $x_{n_k}\in B_{ϵ}\left(y\right),$ and hence belongs to $U.$ But this is our contradiction, because all of the $x_n$ 's were not in $U.$ So, the theorem is proved.