I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
The following exercise is Exercise 4 on p.23 in Exercises 2A in this book.
Exercise 4 on p.23
Suppose $F$ is a subset of $\mathbb{R}$ with the property that every open cover of $F$ has a finite subcover. Prove that $F$ is closed and bounded.
My proof:
The collection $\{ (k,k+2): k\in\mathbb{Z}\}$ is an open cover of $\mathbb{R}$.
So, this collection is also an open cover of $F$.
$F$ has a finite subcover $\{(k_1,k_1+2), (k_2,k_2+2),\dots,(k_n,k_n+2)\}$.
Let $U:=\max\{k_1+2,\dots,k_n+2\}$.
Let $L:=\min\{k_1,\dots,k_n\}$.
Then, $L<x<U$ for any $x\in F$.
So, $F$ is bounded.
Assume that $F$ is not closed.
Then, $\mathbb{R}\setminus F$ is not open.
So, there exists $x_0\in\mathbb{R}\setminus F$ such that $(x_0-\epsilon,x_0+\epsilon)\not\subset\mathbb{R}\setminus F$ for any positive real number $\epsilon$.
So, $(x_0-\epsilon,x_0+\epsilon)$ has an element of $F$ for any positive real number $\epsilon$.
So, $L\leq x_0\leq U$ must hold.
So, $x_0\in (L-1,U+1)$.
There exists $m\in\mathbb{Z}^+$ such that $L-1<x_0-\frac{1}{m}<x_0<x_0+\frac{1}{m}<U+1$.
Then, $$\{(L-1,x_0-\frac{1}{m}),(L-1,x_0-\frac{1}{m+1}),\dots\}\cup\{(x_0+\frac{1}{m},U+1),(x_0+\frac{1}{m+1},U+1),\dots\}$$ is an open cover of $F$.
Since $(x_0-\epsilon,x_0+\epsilon)$ has an element of $F$ for any positive real number $\epsilon$, $F$ doesn't have a finite subcover of the above open cover.
This is a contradiction.
I think my proof is not a standard one.
So, I don't know my proof is ok or not.
Is my proof ok?
Best Answer
In a word, yes.
I feel vaguely that it might be possible to simplify the proof that $F$ is closed, but it's a nice proof. You might want to expand slightly on why your construction using $m$ gives an open cover of $F$: it covers everything in $(L - 1, U + 1)$ except $x_0$, and $x_0 \notin F$.