# Let $X = \Bbb{R}$ with the discrete metric. Is $X$ connected?

No. Any nonempty subset $A ≠ X$ is open, as well as its complement. So $X$ is the union of disjoint nonempty open subsets.

Is there a more formal way of doing it? Thanks for your help.

-
That looks perfectly fine to me. – Nate Eldredge Jun 12 '12 at 5:07
It’s fine. If you wanted to get fancy, you could be specific and let $A=\{0\}$, say, and note that $A$ and $\Bbb R\setminus A$ are open and disjoiont and have $\Bbb R$ as their union. – Brian M. Scott Jun 12 '12 at 5:07
No discrete topology can ever be connected. You answer your own question. What do you mean by "formal way"? – Abhishek Parab Jun 12 '12 at 5:08
@Abhishek: The discrete topology on a singleton is connected... – Brandon Carter Jun 12 '12 at 5:21
@BrandonCarter There are stories of Emil Artin hurling a chalk or duster towards a student who'd ask What about the empty set?" – Abhishek Parab Jun 13 '12 at 3:19

I think your answer is formal enough for any mathematical purpose, but if you want to go fancy you can try the following.

First, prove that a topological space $\,X\,$ is disconnected iff there exists a continuous and onto function $\,f:X\to \{0,1\}\,$ , where the latter space inherits its topology from the usual one on the reals (and, thus, it's a discrete space with two elements).

Now, for your case, show that $\,f:\mathbb{R}_{disc}\to \{0,1\}\,$ defined by $$f(x)=\left\{\begin{array}{ll} 0 \,&\,\text{if}\;x=0\\1\,&\,\text{if}\;x\neq 0\end{array}\right.$$is continuous and onto $\,\{0,1\}\,$...

-
I think writing out arguments like this (which is essentially the same as the proof in the question) only makes a simple fact confusing, is not any more "formal", and certainly not more convincing. – tomasz Jul 13 '12 at 19:29
Perhaps you're right...and that's why I wrote that I think the OP's answer is formal enough mathematically and "...if you want to go fancy...". The point was, of course, to introduce an equivalent definition of disconnected topological space that many times is oversaw. I wouldn't use the above definition for this particular problem, neither. – DonAntonio Jul 13 '12 at 22:36