TITLE: Proof verification: $(0,1)$ is not compact, by definition.
QUESTION [5 upvotes]: I'm toying around with topology, specially about compactness, then I want to know if my proof of the non-compactness of $(0,1)$ is alright.
Consider the topological space $(\mathbb{R},\tau_{e})$, the real line with the euclidean topology.
I want to prove the interval $(0,1)$ is not compact by definition, so it is enough to show an open cover where no finite subcover can exist.
Consider the set of intervals $\left\{ \left(0, 1 - \dfrac{1}{n} \right)\right\}_{n=1}^{\infty}$. I say they form an open cover. I already proved open intervals are open, so it remains to prove they are indeed a cover, i.e., that $(0,1) \subseteq \displaystyle \bigcup_{n=1}^{\infty} \left( 0, 1 - \dfrac{1}{n} \right)$. In fact, take $x \in (0,1)$. Then $x< 1$, so $\dfrac{1}{1-x}>0$, then, by the archimedean property, there is $N$ natural such that $N > \dfrac{1}{1-x}$, so $\dfrac{1}{N} < 1-x$ and then $x < 1 - \dfrac{1}{N}$. So $x \in \left ( 0, 1 - \dfrac{1}{N} \right)$, and of course this is one of the intervals in the set of intervals, hence the set is a cover.
Now I say that it is impossible to extract a finite subcover, because if $M$ is a natural such that $(0,1) \subseteq \displaystyle\bigcup_{n=1}^{M} \left(0, 1 - \dfrac{1}{n} \right) =\emptyset \cup\left(0,\dfrac12 \right) \cup \dots \left(0, 1 - \dfrac{1}{M} \right)$, then take $x$ such that $1 - \dfrac{1}{M}<x<1$ (which can be taken by the density of real numbers), so $x$ is an element of $(0,1)$ that is not in the finite subcover, hence the impossibility. QED

REPLY [3 votes]: What you have done is absolutely correct.
Here is another way which might be useful to you for proving open sets in $\mathbb{R}$ or rather $\mathbb{R}^{n}$ is not compact .

A compact subset $K$ of a Hausdorff space is closed. So it also suffices to show that $(0,1)$ is not closed.

Fix $x\in X\setminus K$. and a point $p\in K$. Then by Hausdorffness we have $U_{x,p},V_{x,p}$ disjoint such that $x\in U_{x,p}$ and $p\in V_{x,p}$ .
Then consider the open cover of $K$ by $\{V_{x,p}\}_{p\in K}$. Then as $K$ is compact ,we have a finite subcover say $\{V_{x,p_{i}}\}_{i=1}^{n}$ . Then correspondingly we have sets $\{U_{x,p_{i}}\}_{i=1}^{n}$. Then take $U_{x}=\bigcap_{i=1}^{n}U_{x,p_{i}}$ . Then $U_{x}$ is a open nbd of $x$ such that $U_{x}\cap K=\phi$.
Then $X\setminus K= \bigcup_{x\in X\setminus K} U_{x}$ is open as it is a union of open sets. So $K$ is closed .