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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.

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    Disclaimer: I remember nothing about metric spaces and topology. As a student, what really helped me with "formalizing proofs" was just using definitions as often as possible. So, for instance, your claim that "any nonempty subset $A \neq X$ is open", well you should prove it by using the definition of an open subset of the real number metric space. Your professor/teacher probably won't care, but it can help you if you think you are being "informal" (aka skipping steps that need to be proved).2012-11-17

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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 $\,\{0,1\}\,$...

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    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.2012-07-13