$R^{3}$ is open and closed I have thought about 2 ways to prove this, but there are not complete. What should I add to finish the proof?
Proof 1: For all $x\in \mathbb{R^{3}}$ there is $r>0$ such that $B(x,r)\subset \mathbb{R^{3}}$
any $r>0$ will satisfy the requirement because $\mathbb{R^{3}}$ is dense? which arguments about $\mathbb{R^{3}}$ can I use.
And therefore $\mathbb{R^{3}}$ is open
On the other hand A=$\mathbb{R^{3}}\setminus \mathbb{R^{3}}=\emptyset$ and the empty set is open and closed so $A$ is open (and closed?)
Proof 2: Every limit point of $\mathbb{R}^{3}$ are contained in $\mathbb{R}^{3}$ so $\mathbb{R^{3}}$ is closed.
which facts on the limit points can I use to prove it is open?

Take $x\in\mathbb{R}^{3}$. Then, for all $\varepsilon >0$, $B_{\varepsilon}(x)\subset\mathbb{R}^{3}$ By contradiction. Suposse that exist $x\in B_{\varepsilon}(x)$ such that $x\notin\mathbb{R}^{3}$, then, $x\in{\mathbb{R}^{3}\setminus\mathbb{R}^{3}}=\emptyset$. Clearly this is a contradiction.
Now, since $\mathbb{R}^{3}$ is open, $\emptyset$ is closed.
In the other way, we want to prove that $\emptyset$ is open, i.e., if a $x\in\emptyset$, then exist $\delta>0$ such that $B_{\delta}{x}\subset\emptyset$. This is true by emptiness. Then, $\emptyset$ is open and his complemente $\mathbb{R}^{3}$ is closed.
