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Let $\tau $ be the topology on $\Bbb R$ for which the intervals $[a,b)$ form a base.Let $\sigma$ be a topology on $\Bbb R$ such that $\sigma \supseteq \tau. $

Prove that if $x\mapsto -x$ is continuous then $\sigma$ is the discrete topology.

In order to show that $\sigma$ is the discrete topology we have to show that each singleton $\{x\}$ is open in $(\Bbb R,\sigma)$ .

Please give some hints on how to show that.

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    $\begingroup$ Hint: $(a-1,a]\cap [a,a+1)=\{a\}$. $\endgroup$ – BigbearZzz May 1 '16 at 4:58
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$i(x) = -x$ is its own inverse and for every $x$ we have $(x-1, x] = i^{-1}[[-x, -x+1) ]$, so this is open in any topology like $\sigma$ which makes it continuous.

So is $[x,x+1)$ by virtue of $\tau \subseteq \sigma$, for any $x$.

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  • $\begingroup$ Okay I got your point $\endgroup$ – Learnmore May 1 '16 at 5:41

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