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 ?

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

Neither $0$ nor $1$ belong to any of the sets $U_n$. It is not a cover of $[0,1]$.

  • $\begingroup$ Can you show be a cover of $[0,1]$ ? $\endgroup$ – Alan Simonin Mar 10 '15 at 21:03
  • 3
    $\begingroup$ Sure. $(-1,2)$ is a cover of $[0,1]$. $\endgroup$ – Umberto P. Mar 10 '15 at 21:04
  • 2
    $\begingroup$ @AlanSimonin what you want is a family of opens such that $\bigcup_n U_n \supseteq [0,1]$ $\endgroup$ – Omnomnomnom Mar 10 '15 at 21:07
  • 2
    $\begingroup$ That's not it. It says that any open cover, no matter what, can be reduced to a finite subcover. $\endgroup$ – Umberto P. Mar 10 '15 at 21:08
  • 2
    $\begingroup$ @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 $U \subset \Bbb R$. $\endgroup$ – Omnomnomnom Mar 10 '15 at 21:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.