Show that unions of circles and $(0,0)$ is compact

Question

Let $C_n$ be the circle defined as $x^2 + y^2 = \frac{1}{n^2}$. Let $$X = \{(0,0)\}\cup\bigcup_{n\in\mathbb{N}} C_n $$

Show that $X$ is compact.

$X$ is obviously bounded, but how should I show that it is closed? It seems clear that it is, but I'm not sure how I can prove it? Maybe I should consider a continuous function from some compact set that maps to $X$? One possible map would be from $(r,\theta)$ where $r\in\mathbb{N}$ and $\theta\in [0,2\pi]$ but it's not clear that the map is continuous...

Answer 1

Take any sequence $x_k$ in $X$. Then either infinitely many of the $x_k's$ are in one of the $C_n's$ or not. If so, then there is a subsequence of $x_k$ that converges to a point in $C_n$. Otherwise, there is a subsequence $x_{n_k}$ such that $x_{n_k}\in C_{\phi(k)}$ for some monotone increasing function $\phi$. In this case, $x_{n_k}\rightarrow 0\in X$.

Answer 2

Hint: Take the function $f(x,y)= \sqrt{x^2+y^2}$, and then show that the set $\{0\}\cup_{n\geq 1} \{\frac{1}{n}\}$ is closed.

Answer 3

If $U = \{U_i\}_{i \in I}$ is an open cover of $X$, let $U_0 \in U$ contain $(0,0)$. Then for all but finitely many $n$, $C_n \subset U_0$. If $R$ is the union of all $C_n$ that are not contained in $U_0$, then as the union of a finite number of compact subspaces $R$ is compact, and there exists a finite subcover of $U$, $U_1 \cup ... \cup U_m \supseteq R$. It follows that $U_0 \cup U_1 \cup ... \cup U_m$ is a finite subcover of $U$ containing $X$. Since $U$ was arbitrary $X$ is compact.