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Suppose that $ \tau $ is a topology on $ \mathbb{R}$ that contains all closed intervals. Prove that $ \tau $ is the discrete topology on $ \mathbb{R}$.

Answer:

Since $ \tau $ contains all closed intervals, all the intervals $ [a,b] , \ \ a,b \in \mathbb{R} $ are open set in $ \tau $. Let $ \epsilon >0 $ be a real number. Then $ [x-\epsilon, x] \cap [x,x+1]=\{x\} \in \tau $ for all x.

So every singleton is open in $ \tau $.

So $ \tau $ is discrete .

Am I right ?

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  • $\begingroup$ I think you should show that the discrete topology also gives rise to the 'closed interval' topology to show that they're equivalent. $\endgroup$ – Osama Ghani Jul 5 '17 at 5:42
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    $\begingroup$ Since the closed interval [x,x] = {x}, singletons are open. $\endgroup$ – William Elliot Jul 5 '17 at 8:15
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    $\begingroup$ You are right but you don't need $\epsilon$ because $[-1+x]\cap [x,x+1]=\{x\}$. $\endgroup$ – DanielWainfleet Jul 5 '17 at 20:13
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Yes. ${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$

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  • $\begingroup$ please mark or vote up my question $\endgroup$ – M. A. SARKAR Jul 5 '17 at 6:05
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    $\begingroup$ I think the answer is too verbose. $\endgroup$ – copper.hat Jul 5 '17 at 6:21
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    $\begingroup$ @copper.hat . I saw a good one-word answer to the Q "Alternatives to math?" on this site. $\endgroup$ – DanielWainfleet Jul 5 '17 at 20:19

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