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For a topological space $(X,T)$ with a basis $B$, is every basis element of $B$ an open set of $X$ (i.e. in $T$)?

(Forgive me for the dumb question, just trying to learn the basics)

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yes, $B$ is a subset of $T$. – Damien L Feb 6 '13 at 0:52
up vote 1 down vote accepted

By definition $B$ is a base for $T$ if and only if $T$ is the set of all unions of subsets of $B$. For any $b\in B$, $\{b\}$ is a subset of $B$, so its union is in $T$. But its union is just the set $b$, so $b\in T$. Thus, $B\subseteq T$.

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One last question...I'm trying to prove that the lower limit topology of the reals (T') is strictly finer than the standard topology on R (T), and first, I want to show that I can pick any element of T and that is in T'. So does picking an open interval (a,b) basis count as picking any element of T? Or are there elements in T that aren't in the basis for R? – user39794 Feb 6 '13 at 1:02
@Allison: There are many $T$-open sets in $\Bbb R$ besides the open intervals: $(0,1)\cup(2,3)$ is a very simple example. However, you can show that every $(a,b)$ with $a<b$ is in $T'$ by showing that it’s a union of sets of the form $[x,b)$, and it will then follow that every $U\in T$ is in $T'$: it’s a union of open intervals, they’re in $T'$, and topologies are by definition closed under taking unions. – Brian M. Scott Feb 6 '13 at 1:08
Thank you for the help!!! – user39794 Feb 6 '13 at 1:12
@Allison: You’re welcome! – Brian M. Scott Feb 6 '13 at 1:15

(just so the question does not remain unanswered) By definition, any arbitrary union has to be in $T$ hence also a `union' with one element.

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