# Compactness of $[0,1]$

Thanks to Heine-Borel Theorem, we know that $[0,1]$ is compact in $\mathbb R$. However, we I try to use the general topology definition of a compact set, I cannot "see" why it is true.

Here is what I have done :

Let $U_n = (\frac1n, 1-\frac1n)$ be a family of open sets of $\mathbb R$. This family is a cover of $[0,1]$ since $\bigcup_{n\in\mathbb N}U_n = [0,1]$. Then by definition, it must exist a finite sub-family $(U_j)_{j\subset \mathbb N}$ with j finite that covers $[0,1]$.

I can't think of one.

Here is my conclusion :

• Either $U_n$ is not an open cover of $[0,1]$ and is an open cover of $(0,1)$.
• Either I didn't understand the definition.
• Either I didn't manage to see a finite subcover.

What do you think ?

• Error in your infinite union. It doesn't capture the endpoints. Mar 10 '15 at 21:03
• If you have a proof of hiene borel handy, you might want to follow it. Mar 10 '15 at 21:08

Neither $0$ nor $1$ belong to any of the sets $U_n$. It is not a cover of $[0,1]$.
• Can you show be a cover of $[0,1]$ ? Mar 10 '15 at 21:03
• Sure. $(-1,2)$ is a cover of $[0,1]$. Mar 10 '15 at 21:04
• @AlanSimonin what you want is a family of opens such that $\bigcup_n U_n \supseteq [0,1]$ Mar 10 '15 at 21:07
• @AlanSimonin that's for a compact space. If we're to consider $[0,1]$ as a space, then the sets $[0,a)$ and $(a,1]$ are open subsets. If we want to show that $[0,1]$ is a compact subset, we want the condition that I describe. The conditions are equivalent since a set in the space $[0,1]$ is open exactly when it can be written as $U \cap [0,1]$ for some open $U \subset \Bbb R$. Mar 10 '15 at 21:14