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Quick question. Say we are given the unit circle $\{ (x,y)\in \mathbb{R}^2: x^2+y^2=1 \}$.

Is this set compact? How can I prove that this is closed? Bounded? Do I have to take the complement of the set, showing that that set is open (and so unit circle is closed)? Any other trick?

In addition, how can I show that $\{(x,y) \in \mathbb{R}^2: x^2+y^2 < 1\}$ is not compact? I have to show that this thing is open, how can I do that?

I know that compact is equivalent by saying that the set is bounded and closed, if we are talking about subsets of $\mathbb{R}^n$. I also can see that the unit discs are bounded, because the distance between any two points in the set is bounded. But how to show that those are open/closed?

Thanks for your help! :-)

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For each point $p$ not on the circle find an open ball centred at $p$ that misses the circle. For each point $p$ in the open disk find an open ball centred at $p$ that lies inside the open disk; you’ll already have done this if you do the other problem first. – Brian M. Scott Mar 21 '13 at 20:45
Actually $\{(x,y) \in \mathbb{R}^2 : x^2 + y^2 \lt 1 \}$ is a basic open set for the metric topology. – hardmath Mar 21 '13 at 20:47
You could also try to show that $S^1$ is the preimage of some closed set under the norm $\|\cdot\|:\mathbb R^2\to \mathbb R$ – Stefan Hamcke Mar 21 '13 at 20:49
@hardmath: sigh Hardway“R”Us. – Brian M. Scott Mar 21 '13 at 20:51
up vote 15 down vote accepted

The set $\{1\} \subset \Bbb R$ is closed, and the map $$f: \Bbb R^2 \longrightarrow \Bbb R,$$ $$(x, y) \mapsto x^2 + y^2$$ is continuous. Therefore the circle $$\{(x,y) \in \Bbb R^2 : x^2 + y^2 = 1\} = f^{-1}(\{1\})$$ is closed in $\Bbb R^2$.

Your set is also bounded, since, for example, it is contained within the ball of radius $2$ centered at the origin of $\Bbb R^2$ (in the standard topology of $\Bbb R^2$).

Since $\{(x,y) \in \Bbb R^2 : x^2 + y^2 = 1\}$ is a closed and bounded subset of $\Bbb R^2$, the Heine-Borel theorem implies that it is compact.

To see that $B = \{(x,y) \in \Bbb R^2 : x^2 + y^2 < 1\}$ is not compact, note that the sequence $x_n = (0, 1 - \tfrac{1}{n})$ in $B$ converges to $(0, 1) \notin B$. Therefore $B$ is not closed. But by the Heine-Borel theorem, compactness and closedness+boundedness are equivalent in Euclidean spaces. Since $\{(x,y) \in \Bbb R^2 : x^2 + y^2 < 1\} \subset \Bbb R^2$ is not closed it cannot be compact.

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This is a lovely argument for showing that the circle is closed! – Tom Oldfield Mar 21 '13 at 20:52
Regarding your last paragraph: being open doesn't imply not being closed. $\mathbb R^2$ is both open and closed. – Cantor Mar 21 '13 at 21:03
@TomOldfield Would you really consider proving the circle is closed otherwise? – 1015 Mar 21 '13 at 22:23
@Cantor: I was a fool. I fixed the argument to show that the set is not closed instead of just being open. Thanks. – Henry T. Horton Mar 21 '13 at 22:35
How are we to understand $f^{-1}$ when $f$ is not one to one? "$f^{-1}$" as defined above is not a function, so how is it a continuous function? $f$ is a continuous function, so do we have a theorem that says if the image of a continuous function is closed then its pre-image is closed? Or am I totally missing something? – Todd Wilcox Mar 22 '13 at 10:14

Let $$f: \Bbb R \longrightarrow \Bbb R^2,$$ $$\theta \mapsto (\cos\theta,\sin\theta),$$ then $f$ is continuous, and the unit circle is $f([0,2\pi]$) and so it's a compact set of $\Bbb R^2$ as image of the compact $[0,2\pi]$ by the continuous function $f$.

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That's the most elementary way, +1. – 1015 Mar 21 '13 at 22:49

Hint: One way to do this is to note that the continuous image of a compact set is compact (Why?)

So to show that the unit circle is compact, you can find some continuous $f:[0,1] \rightarrow C$. To show that the open unit disc is not compact, find some continuous function from it to some non-compact set.

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Closed: Let us take any sequence $\,\{(x_n,y_n)\}_{n\in\Bbb N}\subset S^1:=$ the unit circle, then:

$$\{x_n\}\,,\,\{y_n\}\,\,\,\text{are bounded infinite sequences in}\,\,\Bbb R$$

Apply now the Bolzano-Weierstrass theorem to each of these two sequences.

Boundedness is trivial.

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Sorry to interfere, but these hints seem strange to me. Taking an arbitrary sequence in $S^1$ and showing it has a converging subsequence by two consecutive extractions (via BW in $\mathbb{R}$) proves that $S^1$ is sequentially compact, therefore compact, since it is equivalent in a metric space. So you prove compact all at once, without refering to Heine-Borel. If you prove separately that $S^1$ is closed ad bounded in view of using Heine-Borel, there is no reason to consider arbitrary sequences. – 1015 Mar 21 '13 at 22:16
I wasn't even thinking of sequentally or whatever compact directly, but of proving directly (again) that $\,S^1\,$ is closed by means of showing it contains its limits points... – DonAntonio Mar 21 '13 at 22:24
Well, then you don't start with an arbitrary sequence in $S^1$. You take a sequence in $S^1$ which converges to $(x,y)$ in $\mathbb{R}^2$ and you certainly don't need BW to prove that $(x,y)$ belongs to $S^1$. – 1015 Mar 21 '13 at 22:26
This way, that way...: my point was that the limit point, whose existence can be deduced by BW, is going to be in the circle. – DonAntonio Mar 21 '13 at 22:42
I highly respect you and all the great answers you have produced here. That being said, this one is problematic. If you want to prove a limit point belongs to something, you take it first...You don't prove its existence... – 1015 Mar 21 '13 at 22:45

Let $|| \cdot||$ denotes the euclidean norm on $\mathbb{R}^2$, there exists a characterization of a compact set in $\mathbb{R}^n$ saying that:

$E$, a non-void subset of $\mathbb{R}^n$, is compact if and only if for every sequence $(x_{n})_{n \ge 1}^{\infty}$ $\subset E$ there exists a sub-sequence $(x_{n_{k}})_{k \ge 1}^{\infty}$ that converges in $E$ (i.e: $\exists$ $x^{*}$ $\in$ $E$ such that $lim_{k \rightarrow \infty}$ $x_{n_{k}} = x^{*}$).

Therefore in that case if we consider that $E =$ the unit circle and, given a sequence $(x_{n})_{n \ge 1}^{\infty}$ $\subset E$, we can see that $\forall x \in E, ||x||=1$, So we can conclude that $E$ is bounded ($||x|| < 2$, $ \forall x \in E $ for example). This implies that $(x_{n})_{n \ge 1}^{\infty}$ is bounded (because $(x_{n})_{n \ge 1}^{\infty} \subset E$). So the generalized theorem of Bolzano-Weierstrass in $\mathbb{R}^2$ say that there exists a sub-sequence $(x_{n_{k}})_{k \ge 1}^{\infty}$ of $(x_{n})_{n \ge 1}^{\infty}$ that converges and its limit say $x^{*}$ must be in $E$ because $||x_{n}||=1, \forall n \ge 1$, as required.

Now the subset $F=\{(x,y)\in \mathbb{R}^2: x^{2}+y^{2} < 1 \}$ denoted by $B((0,0),1)$ and called the open ball centered in $(0,0)$ of radius $1$ is an open-set because given $x \in F$ if we consider the open-ball $B(x, \delta)$ with $\delta > 0$ such that $\delta < (1-||x||)$, we easily see that $B(x,\delta) \subset F$ And so $F$ is an open space according to the definiton of an open-set in $\mathbb{R}^2$, so $F$ can't be closed (because $\mathbb{R}^2$ is a connected space and thus the only subsets of $\mathbb{R}^2$ that are at the same time open and closed are $\emptyset$ and $\mathbb{R}^2$...). Finally, we conclude that $F$ is not compact. (Because F is not closed, even if it's bounded).

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