I have a quick question
In $\mathbb{R}$ let $$S \subset \mathbb{R}$$
If S is not open then must it be a closed set?
I think no because if it not closed means its complement is open. But how do I show this?
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Show this by exhibiting a set that is neither open nor closed. – BrianO Oct 30 '15 at 20:28
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2The set $[0,1)$ is neither open nor closed. – Stefan Hamcke Oct 30 '15 at 20:28
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You can show a statement is false by exhibiting a single counterexample. – MPW Oct 30 '15 at 20:29
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What if I have if S is not closed then must it be open? – Fernando Martinez Oct 30 '15 at 20:32
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if for all S, "S not closed $\to$ S open", then by taking complements "S not open $\to$ S closed"... so, do the logic. – BrianO Oct 30 '15 at 20:40
1 Answers
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In $\mathbb R$ the only subsets which are open and closed at the same time are $\varnothing$ and $\mathbb R$. This is due to the fact that $\mathbb R$ is connected.
To show that, say $\varnothing \subsetneq A\subsetneq \mathbb R$, and that $A$ is both open and closed, and say $x\in A$, $y\not\in A$, and assume that $x<y$. Then defined $$ s=\sup\{t\in A : t<y\}. $$ If $t\in A$, then $(t-\varepsilon,t+\varepsilon)\subset A$, for some $\varepsilon>0$, as $A$ is open, and hence $t$ is not the supremum of the set above, a contradiction. If $t\in \mathbb R\smallsetminus A$, then as $\mathbb R\smallsetminus A$ is open, there exists an $\varepsilon>0$, such that $(t-\varepsilon,t+\varepsilon)\subset \mathbb R\smallsetminus A$. Again leads to a contradiction, as in such case $\sup\{t\in A : t<y\}\le t-\varepsilon$.
Yiorgos S. Smyrlis
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