How to determine the closure of a subset and prove it is actually the closure? I have this subset $E = \{r \in Q: r^2 \leq 2\}$ which is in $\mathbb{R}$ with the Euclidian metric. I was wondering how can I find the closure of this subset. Here is what I have:
The limit points of the subset is the closed interval $[-\sqrt{2}, \sqrt{2}]$, thus this is the closure. Is this right? If so can someone help me to prove it? How would one prove closure of a subset?
Thank you.
 A: First note that in order to prove that a set $B$ is the closure of a set $A$ it suffices to prove that the intersection of any open ball of some arbitrary $p\in B$ with $A$ is not empty (you can prove that a set $A$ is closed iff it is equal to its closure). 
Now, let $x$ be any point in the complement of $[-\sqrt2, \sqrt2]=E$ and define $r:=\min\{|x-\sqrt2|,|x+\sqrt2|\}.$ What can you say about the open ball $B_r(x)$ (radius $r$ and center at $x$)? 
A: Jup, you are right! Let $\alpha$ be in $I=]-\sqrt{2},\sqrt{2}[$. Then, there exists a sequence $(x_{n})$ in $\mathbb{Q}$ converging to $\alpha$. So you can find $N \in \mathbb{N}$ such that $n \geq N \Rightarrow |x_{n}-\alpha|<\sqrt{2}-|\alpha|$, so for $n\geq N$, $|x_{n}| \leq |\alpha|+|x_{n}-\alpha|<\sqrt{2}$. Define $(y_{n})=x_{n+N}$ and you get a sequence of rationals in E converging to any element $\alpha$. So $I \subseteq CL(E)$ (I will leave the boundaries for you, it's not really hard, you can  follow almost the same steps!).Now let $(x_{n})$ be a sequence in $E$ converging to $x \in \mathbb{R}$. Then by comparaison, $x \in [-\sqrt{2},\sqrt{2}]$, since $-\sqrt{2} < x_{n} <\sqrt{2}$.QED
A: Prove it the same way you prove that any two sets are equal:

$$ A = B \;\; \text{iff} \; \; A \subseteq B \text{ and } B \subseteq A$$

Inclusion 1: Let $x \in [-\sqrt{2},\sqrt{2}]$. If $x \in \mathbb{Q}$, we are done. If not, then there is a sequence of rationals $\{x_n\}$ such that $x_n \to x$ (justify this either by the definition of irrational numbers or that rationals are dense in the reals). We can truncate this sequence to include only rationals in $[-\sqrt{2},\sqrt{2}]$. But this is the very condition of being a limit point, so $x \in E'$.
Inclusion 2: Let $x \in E \cup E'$. If $x \in E$, then we immediately have $x \in [-\sqrt{2},\sqrt{2}]$. If $x \in E'$, then there is a sequence $\{x_n\} \subset E$ such that $x_n \to x$. Since $-\sqrt{2} < x_n < \sqrt{2}$, we have $-\sqrt{2} \leq x \leq \sqrt{2}$ (if this isn't obvious, it is a simple proof by contradiction). In either case, $x \in [-\sqrt{2},\sqrt{2}]$.
