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To show that the only nonempty subset of $\Bbb R$ which is both open and closed in $\Bbb R$ is $\Bbb R$.

Let $A$ be a non empty subset of $\Bbb R$ which is both open and closed. Let $x \in A$. Then $(x- \epsilon ,x+ \epsilon) \subset A$ for some $\epsilon >0$. Then how to bring the contradiction by putting in the fact that $A$ cannot be simultaneously open and closed!!

User8976
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    This is kind of difficult to prove directly from first principles. The proof is that $\mathbb{R}$ is connected, so such a thing is impossible. $\mathbb{R}$ is connected since it's homeomorphic to an open interval, which is connected. This is proved using the least upper bound property of the ordering on $\mathbb{R}$, and the proof is a bit involved. See Munkres section 24. – Moya Mar 06 '15 at 06:04
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    related: http://math.stackexchange.com/questions/70415/showing-that-mathbbr-is-connected – Closure Mar 06 '15 at 06:05

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If $A$ is nonempty, closed, and open (clopen), then $A$ and $\mathbb{R}-A$ are both open. Therefore, the pair $$A,\mathbb{R}-A$$

is a separation of $\mathbb{R}$, which contradicts the fact that $\mathbb{R}$ is connected. To see that $\mathbb{R}$ is connected, you can check Showing that $\mathbb{R}$ is connected.

Community
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The Substitute
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    This is more like a link than an answer? –  Mar 06 '15 at 06:45
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    Not really, it's an answer that uses a simple fact. A link was provided only if that fact needed to be proved seperately – Alan Mar 06 '15 at 07:28