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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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    That looks perfectly fine to me.2012-06-12
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    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.2012-06-12
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    No discrete topology can ever be connected. You answer your own question. What do you mean by "formal way"?2012-06-12
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    This looks formal enough. If you want to chain things together: (1) The discrete metric gives a space the discrete topology (2) The discrete topology is connected if and only if the underlying set has cardinality $\leq 1$.2012-06-12
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    @Abhishek: The discrete topology on a singleton is connected...2012-06-12
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    @Dylan: Incidentally, many prefer the convention that the empty set is not connected. (I don't really know why.)2012-06-12
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    @Jonas I worried about that too, but the Wikipedia page didn't seem to care. I don't consider the empty set to be an irreducible topological space, though...2012-06-12
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    @BrandonCarter There are stories of Emil Artin hurling a chalk or duster towards a student who'd ask ``What about the empty set?"2012-06-13
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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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