Show that {$\sqrt 2$} $\cup$ {$\sqrt 2$ + $1/n$ : $n$ $∈$ $\mathbb{N}$} is closed I want to show that {$\sqrt 2$} $\cup$ {$\sqrt 2$ + $1/n$ : $n$ $∈$ $\mathbb{N}$} is closed. I'm having trouble. I've been trying to show that the complement is open, but the presence of {$\sqrt 2$} is confusing me.
 A: Let's call your set $A$.

The easiest way to show that $A$ is closed is by seeing that the complement of $A$ is the set
$$(-\infty, \sqrt{2})\cup (\sqrt 2+1,\infty)\cup \bigcup_{n=1}^\infty \left(\sqrt 2+ \frac{1}{n+1}, \sqrt 2 + \frac1n\right)$$
which is pretty clearly open. Of course, you still have to show that what I wrote is in fact the complement of $A$, but that part is also not difficult.

Another simple way to show $A$ is closed is by seeing that the only limit point of $A$ is $\sqrt{2}$, and since $\sqrt2\in A$, $A$ contains all limit points and is closed.
You still have to prove that $\sqrt 2$ is the only limit point, and while this isn't entirely trivial, it is not difficult.

However, if you want to go the long way round, you can take any $x$ from the complement. Then you have three options:


*

*$x<\sqrt{2}$. In this case, you can easily find some $\epsilon$ such that $(x-\epsilon, x+\epsilon)$ does not intersect with your set.

*$x>\sqrt 2 + 1$. Similar case as above.

*$\sqrt 2 < x < \sqrt 2 + 1$. In this case, there exists some $n\in\mathbb N$ such that $\sqrt 2 + \frac 1{n+1} < x < \sqrt 2 + \frac 1n$ (WHY?). Then there exists some $\epsilon$ for which $\sqrt{2}+\frac1{n+1} < x-\epsilon < x+\epsilon < \sqrt 2 + \frac 1n$ (you can calculate it directly from $x$!) and you are done.

A: A set is closed iff it contains all its limit points. The only limit point of the given set is $\sqrt{2}$.
