I have proved the following statement and I would like to know if I have made any mistakes, thank you:
"$F\subset\mathbb{R}$ such that every open cover of $F$ has a finite subcover $\Rightarrow F$ is closed and bounded"
My proof:
$F$ is bounded: For $f\in F$ let $O_1(f)$ be an open interval centered in $f$ of radius $1$: then $\{O_1(f):f\in F\}$ is an open cover of $F$ which by hypothesis has a finite subcover $\{O_1(f_1),\dots, O_N(f_N):f_1,\dots, f_N\in F,\ 1\leq k\leq N\}$. Then if $f_{\max}=\max\{f_1,\dots,f_N\}$ and $f_{\min}=\min\{f_1,\dots,f_N\}$ we have that $f_{\min}-1\leq f\leq f_{\max}+1\ \forall f\in F$ so if we set $M:=\max\{|f_{\max}+1|,|f_{\min}-1|\}$ we have that $F\subseteq B_{M}(0)$ so $F$ is bounded, as desired.
$F$ is closed: Suppose for sake of contradiction $F$ is not closed: then there exists $\bar{x}$ limit point of $F$ such that $\bar{x}\notin F$ so $B_{\frac{1}{n}}(\bar{x})\cap F\neq\emptyset\ \forall n\geq 1$. Consider, for $n\geq 1$, the sets $U_{\frac{1}{n}}:=\{x\in\mathbb{R}: |x-\bar{x}|>\frac{1}{n}\}=(-\infty,\bar{x}-\frac{1}{n})\cup (\bar{x}+\frac{1}{n},+\infty)$, which are all open (since $U_{\frac{1}{n}}^c =[\bar{x}-\frac{1}{n},\bar{x}+\frac{1}{n}],\ n\geq 1$ is closed).
The collection $\{U_{\frac{1}{n}}:n\geq 1\}$ also contains all of $F$ in fact if $f\in F$ then:
- $f\notin B_{\frac{1}{n}}(\bar{x})\cap F,\ n\geq 1\Rightarrow$ either $f<\bar{x}-\frac{1}{n}$ or $f>\bar{x}+\frac{1}{n}$ i.e. $f\in U_{\frac{1}{n}}$;
- $f\in B_{\frac{1}{n}}(\bar{x})\cap F,\ n\geq 1\Rightarrow $ since $\frac{1}{n}\overset{n\to\infty}{\to}0$ there is some $n^*$ such that $f\notin B_{\frac{1}{n^*}}$ so either $f>\bar{x}+\frac{1}{n^*}$ or $f<\bar{x}-\frac{1}{n^*}$ thus $f\in U_{\frac{1}{n^*}}$
Thus $\{U_{\frac{1}{n}}:n\geq 1\}$ is really an open cover of $F$. By hypothesis this open cover admits a finite subcover $\{U_{\frac{1}{n}}:1\leq n\leq N\}$ but if $f\in B_{\frac{1}{N+1}}(\bar{x})\cap F$ then $f\notin \bigcup_{n=1}^{N}U_{\frac{1}{n}}$ so $\{U_{\frac{1}{n}}:1\leq n\leq N\}$ does not cover all of $F$, contradiction. Thus $F$ must be closed, as desired.
Best Answer
Your proof is correct, but the first part ($F$ bounded) is too complicated. Before I explain this point, let us note the condition "Every open cover of $F$ has a finite subcover" can be formulated for any topological space $F$ and says that $F$ is compact.
Now let $(X,d)$ be a metric space. It is a topological space by declaring the open subsets of $X$ to be all possible unions of open balls $B_r(x) = \{ x' \in X \mid d(x',x ) < r\}$ with $r >0$ and $x \in X$. Then $F \subset X$ is called bounded if $F \subset B_R(x_0)$ for some $x_0 \in X$ and some $R = R(x_0) > 0$. Using the triangle inequality, we see that in this case each $x \in X$ admits $R(x) >0$ such that $F \subset B_{R(x)}(x)$. You deal with the special case that $X = \mathbb R$ and $d(x,y) = \lvert x - y \rvert$.
So let $F$ be a compact subset of a metric space $(X,d)$. Pick any $x_0 \in X$. Clearly $F \subset X = \bigcup_{n=1}^\infty B_n(x_0)$. Hence finitely many $B_{n_i}(x_0)$ cover $F$. Let $N =\max{n_i}$. Then $F \subset B_N(x_0)$.