0

If $S$ is a compact subset of $\mathbb{R}$ and $T$ is a closed subset of $S$, then $T$ is compact.

How can I show this using the definition of compactness, and separately showing this by the Heine-Borel theorem.

Trancot
  • 4,021
  • 1
    How can you easily extend an open cover of $T$ to an open cover of $S$? – Daniel Fischer Sep 03 '13 at 19:32
  • Using Heine-Borel would be a bit circular since the fact that closed subsets of compact sets are compact is also used in the proof of Heine-Borel. – Stefan Hamcke Sep 03 '13 at 19:33
  • 1
    Googling "site:math.stackexchange.com closed subset compact" yielded a lot of close-to-duplicate questions. You should try something similar to this if you have a feeling your question might be common. – rschwieb Sep 03 '13 at 19:37
  • I go against the closing votes, because the suggested duplicates mention metric spaces and OP only mentions the real line ; possibly this will make some answers sound a little bit off context for OP if he does not know about metric spaces. – Patrick Da Silva Sep 03 '13 at 19:38

2 Answers2

1

The Heine-Borel theorem states that the compact sets in $\mathbb{R}$ are those closed and bounded. Since $S$ is compact and $T\subset S$ then $T$ is also bounded. A basic result in general topology states the following (Dugundji 1965, p. 78):

Let X be a topological space and $S\subset X$ a closed subspace. Then every closed subset $T$ in the subspace topology relative to $S$ is closed in $X$.

This gives that $T$ is also closed in $\mathbb{R}$ and therefore compact.

Mauricio Tec
  • 2,624
1

People have given the idea for the Heine-Borel part but not for the compactness part.

Given an open cover of $S$, since $S$ is closed, say $S \subseteq \bigcup_{\alpha \in A} \mathcal O_{\alpha}$ where $\mathcal O_{\alpha}$ is open, then $T \subseteq (\mathbb R \backslash S) \cup \bigcup_{\alpha \in A} \mathcal O_{\alpha}$ is an open cover of $T$ (why?). I am letting you finish the proof!

Hope that helps!

Patrick Da Silva
  • 41,413
  • What is $\mathcal{O}$? – Trancot Sep 03 '13 at 19:42
  • It's a letter $O$ but with a mathcal on it (i.e. \ mathcal { O } in math mode). It stands for an open set ; what I did is I wrote an expression for an arbitrary open cover of $S$ ; such a open cover is a family of open sets whose union contain $S$. To denote the open sets in that family, I used an index set $A$ and each open set $\mathcal O$ has its own index $\alpha$. – Patrick Da Silva Sep 03 '13 at 20:05
  • For an example of how you could explicitly use this definition, if for every number $x \in \mathbb R$ you could consider an open set of the form $\mathcal O_x = ]x-\delta_x, x+\delta_x[$ with $\delta_x > 0$, then $\bigcup_{x \in [0,1]} \mathcal O_x$ would be an open cover of $[0,1]$ and $A = [0,1]$ is the set of indices. So you don't always use ${1,2,\dots,n}$ to index your things in mathematics. :) – Patrick Da Silva Sep 03 '13 at 20:08