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
Best Answer
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 .
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 .