# Why is this family of open sets a cover for $(0,1)$?

I'm dealing with some beginner examples of compact spaces. The definition I am given is

A subset $A$ of a topological space $X$ is compact if every open cover for $A$ has a finite subcover for $A$

Then the book shows that $(0,1)$ is not compact. The main idea is that te family of open sets $\{ (1/n, 1): n\in \mathbb{N}, n>1\}$, which covers $(0,1)$ has no single finite subfamily which covers $(0,1)$, because, lets say that the finite subfamily is $\{(1/n_1,1), ... (1/n_r, 1))\}$ covers only $(1/N,1)$, with $N=\max(n_1, n_2, ... , n_r)$.

So I see that this is just applying the definition. But how can I understand that the family of open sets$\{ (1/n, 1): n\in \mathbb{N}, n>1\}$, covers $(0,1)$?

For any $x\in(0,1)$ we need to show there is some $n$ such that $x\in(\frac{1}{n},1)$.

Since $x\neq 0$, we can find such $n$ by finding $n>\frac{1}{x}$ since $x\neq 0$, then $\frac{1}{n}<x$.

You might be confusing "covers" with "covers distinctly." The $x$ can be in more than one of the sets. Indeed, in this case, it is in infinitely many.

Hint: The sequence $(1/n)$ tends to $0$ for $n \to + \infty$. So for $x>0$ and for $n$ large enough, $x>1/n$.

• Yes i know, but if I read that that family 'covers' the set, I immediately think about the 'gaps' in the set I defined... As you can read, I'm very vague, probably beacause I do not deeply understand these things :-) – Applied mathematician Mar 21 '13 at 14:41
• There only 'gap' in your covering set is the one between $0$ and $1/n$, and that gap gets infinitely small. Notice that $(1/2, 1) \subseteq (1/3, 1) \subseteq (1/4, 1)...$ – Tyler Mar 21 '13 at 14:45
• If $x\in(0,1)$, then $x>0$. Since $1/n\to 0$, eventually you find an n so that $x>1/n$, which means $x\in(\frac{1}{n},1)$. – Neal Mar 21 '13 at 14:45
• You don't need $\frac{1}{n}\to 0$ you just need $n>\frac{1}{x}$. That is, every real number has an integer greater than it. – Thomas Andrews Mar 21 '13 at 14:49
• The family $\{ (1/n,1), n \in \mathbb N \}$ covers $(0,1)$ because for all $x \in (0,1)$ there exists a set $(1/n,1)$ that contains $x$. – roger Mar 21 '13 at 14:49

Observe that $0<x<1$ if and only if there exists $n\geq 1$ such that $1/n<x<1$. The "if" is trivial. For the "only if", you can take explicitly $n=\lfloor 1/x\rfloor+1$, as this yields $n>1/x$. Therefore $$(0,1)=\bigcup_{n\geq 1}\left(\frac{1}{n}, 1\right)$$ Note that cover only requires $\subseteq$. But while we're at it...

An easier way of proving $(0, 1)$ is not compact is to show $\mathbb{R}$ is not compact since $(0, 1) \cong \mathbb{R}$ (assuming you've already seen this proof and proved that continuous surjections preserve compactness). The set $\displaystyle \bigcup_{n = 1}^\infty (-n, n)$ is an open cover of $\mathbb{R}$ and does not reduce finitely. Same idea as the original example but a little easier to visualize.