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

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

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 ?

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

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