Exercise on Compact space of R^2 Given $n\in \mathbf{N}$, consider subspace $A_n$ of $\mathbf{R}^2$ defined by
$$A_n:=\left\{(x,y) \in \mathbf{R}^2: x^2+(y-1)^2=\frac{1}{n^2}\right\}$$
and define $\displaystyle X:=\bigcup_{n\in \mathbf{N}} A_n.$
Let 
$$C^+:= \left\{(x,y) \in \mathbf{R}^2:y\geq\frac{1}{2}\right\}$$ and 
$$C^-:= \left\{(x,y) \in \mathbf{R}^2:y\leq\frac{1}{2}\right\}.$$ 

Verify whether $X\cap C^+$ and $X\cap C^-$ are compact topological spaces.

I know that only accumulation point's of $X$ is $(0,1)$, since $X\cap C^+$ and $X\cap C^-$ are limited, I concluded that $X\cap C^+$ is not compact because it is not closed and $X\cap C^-$ is compact because it is closed. Is it right?
 A: I have one minor criticism, but otherwise I agree with your argument (including its conclusion):

I know that the only accumulation point of $X$ is $(0, 1)$

Considered as a subset of $\Bbb{R}^2$, $(0, 1)$ is not the only accumulation point of $X$. Recall that accumulation point $x$ of a set $C$ satisfies $x \in \overline{C \setminus \{ x \}}$. In a metric space like $\Bbb{R}^2$, this is equivalent to the existence of a sequence $x_n \in C \setminus \{x\}$ that converges to $x$. As an example, the points
$$(\cos(1/n), 1+ \sin(1/n))$$
belong to $A_1$, and hence $X$. They converge to the point $(1, 1)$, but no point in the sequence is equal to $(1, 1)$. Thus, $(1, 1)$ is also an accumulation point of $X$ in $\Bbb{R}^2$ (and similar constructions can be used to show the same is true for any point in $X$).

Is $(0,1)$ an accumulation point's of $X \cap C^−$??

No. The point $(0, 1)$ does not lie in $\overline{X \cap C^- \setminus\{(0, 1)\}}$. To prove this, consider an open ball around $(0, 1)$ of radius $\frac{1}{2}$. For $(x, y)$ in this ball, we have
\begin{align*}
\|(x, y) - (0, 1)\| < \frac{1}{2} &\implies |y - 1| \le \sqrt{x^2 + (y - 1)^2} < \frac{1}{2} \\
&\implies -\frac{1}{2} < y - 1 < \frac{1}{2} \\
&\implies \frac{1}{2} < y < \frac{3}{2} \\
&\implies (x, y) \notin C^- \\
&\implies (x, y) \notin X \cap C^- \setminus \{(0, 1)\}.
\end{align*}
So, $(0, 1)$ lies in the interior of the complement of $X \cap C^- \setminus \{(0, 1)\}$, and hence not in $\overline{X \cap C^- \setminus \{(0, 1)\}}$, showing it's not a limit point of $X \cap C^-$.
