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! :-)
$\endgroup$ 45 Answers
$\begingroup$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.
$\endgroup$ 8 $\begingroup$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 $f$.
$\endgroup$ 2 $\begingroup$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.
$\endgroup$ $\begingroup$Hints:
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.
$\endgroup$ 9 $\begingroup$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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