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Consider the closed intervals $[a,b]$ as a sub-basis for a topology on the real line. Describe the resulting topology

My attempt

If $[a,b]$ is open, and $[a-1,a]$ is open, then the intersection of the two = {a} and is also open, since the intersection of open sets is open. You end up with the discrete topology, in which every set is open, since every set is the union of open sets, viz., the sets containing one of its elements.

Is it right to my attempt?, Is there another way to solve it?

Thanks for your help

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If $[a,b]$ is open, and $[a-1,a]$ is open, then the intersection of the two $= {a}$ and is also open ... This looks fine to me, except I would simply observe that for each real number $a$ we have $\{a\}$ open, because $[a-1,a] \cap [a,a+1] = \{a\}.$ That is, I would eliminate the use of the parameter $b$ in the proof, as it's not needed and just adds clutter. –  Dave L. Renfro Jan 23 '13 at 17:58
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