Show this set is closed As part of a proof I am writing for analysis, I need to show the following set is closed:
$F_n = \{x \in \mathbb{R} \, | \,x \ge 0, ~~ 2-\frac{1}{n} \le x^2 \le 2+\frac{1}{n}\}$.
My current approach is to show that $F_n = \overline{F_n}$. I.e. I want to show that an accumulation point $x$ of $F_n$ is in $F_n$. My idea is proof by contradiction, so suppose $x \in (F_n)^c \Rightarrow 2 - \frac{1}{n} > x^2$ or $x^2> 2+\frac{1}{n}$. I try using the fact that $x$ is an accumulation point to reach a contradiction but cannot seem to do so. Is there another approach that is more intuitively obvious? Or am I doing something wrong?
 A: Idea: take any sequence $\;\{x_m\}_{m\in\Bbb N}\subset F_n\;$ . Since it is a bounded sequence, Bolzano-Weierstrass Theorem gives a convergent subsequence $\;\{x_{m_k}\}_{k\in\Bbb N}\;$, say $\;x_{m_k}\xrightarrow[k\to\infty]{}x\;$ , and since  
$$\forall\,k\;,\;\;2-\frac1n\le x_{m_k}^2\le 2+\frac1n\iff \sqrt{2-\frac1n}\le x_{m_k}\le\sqrt{2+\frac1n}\implies$$
$$\implies \sqrt{2-\frac1n}\le x\le\sqrt{2+\frac1n}\implies x\in F_n$$
(Observe that both sides of the double inequality above are constant wrt $\;m\;$ and $\;m_k\;$ !)
A: Fix $n$ and let $x$ be in the adherence of $F_n$. The point $x$ is therefore limit of a sequence $(x_p)_p$ of $F_n$. Each $x_p$ satisfies the inequalities defining $F_n$, so that by passing to the limit as $p$ tends to $+\infty$, the point $x$ satisfies them as well, showing thereby that $x\in F_n$. This shows that the adherence of $F_n$ is included in $F_n$. The reverse inclusion being obvious, the adherence of $F_n$ is equal to $F_n$, and $F_n$ is closed indeed.
A: the square root is continuous so the inverse image of $[2-\frac1n,2+\frac1n]$ is closed. 
