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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


(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.

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up vote 3 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.

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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. – T. Eskin 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. – T. Eskin Feb 25 '13 at 13:58
@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 $\mathbb{R}$. – T. Eskin Feb 25 '13 at 15:57

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