Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm working on compactness in topological spaces and I wanted to check I am correctly understanding and implementing some theorems.

So taking a common example, is

$S =\{ (x,y) \in \mathbb{R}^2 : x^2 + y^2 =1 \}$ compact in $\mathbb{R}^2$?

As stated in Sutherland's Introduction to Metric and Topological spaces p136 Exercise 13.4. (It doesn't state that topology being used so I will assume the standard Euclidean).

This is the way I have approached the problem: Using the theorems that

Any closed subset of a compact set is compact

and

(Heine-Borel) Any closed bounded subset of $\mathbb{R}^n$ is compact

I have concluded that $C = [-1,1] \times [-1,1] \subset \mathbb{R}^2$ is compact and as the complement of $S$ in $C$ is open, $S$ is closed and thus compact in $C$.

However I run into problems here because I have compactness in C and not $\mathbb{R}^2$, and I have a strong feeling that compactness in a set doesn't necessarily imply compactness in the superset.

I would really appreciate your help in understanding the best way to approach this type of problem so I can feel confident in more complex examples. Thanks in advance.

P.S Apologies if I have formatted or referenced anything incorrectly, I'm still very new here.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Compactness is a topological property that does not depend on the superset, so to speak. If $X$ is a topological space and $Y\subseteq X$, then $Y$ is a compact subset of $X$ if and only if $Y$ is compact in its subspace topology (as a topological space).

To prove this, assume that $Y$ is a compact subset of $X$ and let $\{U_{i}\}_{i\in I}$ be an open cover of $Y$ in its subspace topology. By definition, for each $i$ there exists an open $V_{i}\subseteq X$ with $U_{i}=V_{i}\cap Y$. Now $\{V_{i}\}_{i\in I}$ is an open cover of $Y$ in $X$. By compactness of $Y$ we may extract a finite subcover $\{V_{i_{k}}\}_{k=1}^{n}$, and now $\{U_{i_{k}}\}_{k=1}^{n}$ is the finite subcover of $\{U_{i}\}_{i\in I}$ in the subspace topology that we were looking for. Conversely, assume that $Y$ is a compact topological space and $Y\subseteq X$, where $Y$ has the subspace topology of $X$. Take an open cover $\{U_{i}\}_{i\in I}$ of $Y$ in $X$. Now $V_{i}:=U_{i}\cap Y$ is open in $Y$ for each $i$, and $\{V_{i}\}_{i\in I}$ is an open cover of $Y$ in its subspace topology. Use compactness to extract a finite subcover $\{V_{i_{k}}\}_{k=1}^{n}$, whence $\{U_{i_{k}}\}_{k=1}^{n}$ is the desired finite subcover of $Y$ in $X$.

Another quick approach to your problem: note that the function $f:\mathbb{R}^{2}\to\mathbb{R}$, $(x,y)\mapsto x^{2}+y^{2}$ is continuous, and $S=f^{-1}\{1\}$. Since diam$(S)=2$, then $S$ is bounded. Apply Heine-Borel.

share|improve this answer
    
Your quick approach may suggest that any closed and bounded set in any metric space is compact, which is not true of course. –  Asaf Karagila Feb 25 '13 at 13:38
    
@AsafKaragila: For this reason I said to apply Heine-Borel :) By this I mean the theorem which states that a subset of $\mathbb{R}^{n}$ is compact iff it is bounded and closed. In general, a subset of any metric space is compact iff it is totally bounded and complete. Also (as a side note) one may note that completeness is equivalent with closedness in $\mathbb{R}^{n}$ and total boundedness with boundedness. –  Thomas E. Feb 25 '13 at 13:40
    
@ThomasE. Thank you for your quick response, I was sure there was a quick method which was eluding me. –  120 Eyes Feb 25 '13 at 13:47
    
@120Eyes. Sure. See also my edit for your question about the compactness. –  Thomas E. Feb 25 '13 at 13:58
1  
@120Eyes. A function $f:X\to Y$ is continuous if and only if $f^{-1}F$ is a closed subset of $X$ for every closed subset $F$ of $Y$. The singleton $\{1\}$ is a closed subset of $\mathbb{R}$. –  Thomas E. Feb 25 '13 at 15:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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