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I'm trying to understand a pretty simple theorem in my intro to Topology notes:

Let $X$ be a topological space and $A \subset X$ . Suppose the topology is given by a basis $B'$; then $x$ is in the closure of $A$ iff every basis element $B$ containing $x$ intersects $A$.

Before this theorem in my notes the author states:

$x \in $ cl($A$) iff $A$ intersects every open set $U$ containing $x$.

First I have a preliminary question, is the basis of a topology necessarily contained in the topology? Ex.: if not, the interval (in this case our topology with open sets in $(0,1)$)

$(0,1)$ could have as a basis all open intervals in $\Bbb R$ (?)

Secondly, my notes state for the forward direction: If every open set containing $x$ intersects $A$, then every basis element containing $x$ intersects $A$. (In my notes "intersect" means the intersection isn't empty). I think this statement is supposed to be obvious, but it is not obvious to me at all. Any insights appreciated. Also I am new to this topic so if anything I've written doesn't make sense I will elaborate or clean it up immediately.

IntegrateThis
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  • The definition of a base (basis) for a topology $T$ is a set $B$ such that $T={\cup C: C\subset B}.$ That is, $ t\in T$ iff $t$ is the union of the members of some $C\subset B.$... Members of $B$ are members of $T$ because if $b\in B $ and $C={b}$ then $C\subset B$ so $b=\cup C\in T.$... Note that for a given $t\in T$ the set $C$ need not be unique... Bases are useful tools. – DanielWainfleet Sep 12 '18 at 01:58
  • $T$ itself is a base for the topology $T.$.... The set $B$ of bounded open real intervals with rational end-points is a "nice" base for the usual topology on $\Bbb R$ because there are uncountably many open subsets of $\Bbb R$ but $B$ is countable. – DanielWainfleet Sep 12 '18 at 02:04

1 Answers1

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Every basis element is open. If $x \in \mathrm{cl}(A)$ then every open set containing $x$ intersects $A$, so in particular every basis element containing $x$ intersects $A$.

On the other hand, suppose every basis element containing $x$ intersects $A$. If $O$ is an open set containing $x$, there exists a basis element $B$ with $x \in B \subset O$. Since $B$ intersects $A$, so does $O$. Thus every open set $O$ containing $x$ intersects $A$, so that $x \in \mathrm{cl}(A)$.

Umberto P.
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  • My notes don't say the basis elements are all open in the definition of a basis. This wolfram definition is the same as mine : http://mathworld.wolfram.com/TopologicalBasis.html – IntegrateThis Sep 12 '18 at 00:05
  • @IntegrateThis, it's not part of the definition, it just follows immediately from the definition. Everything in the basis can (trivially) be written as a union or finite intersection of things in the basis. –  Sep 12 '18 at 00:13
  • By the way, it should say "union of finite intersections" not or. – GEdgar Sep 12 '18 at 00:14
  • Ok. I understand now, thanks for the help. For anyone reading this in the future I found this helpful as well :https://math.stackexchange.com/questions/2248800/are-basis-elements-of-a-topology-always-open – IntegrateThis Sep 12 '18 at 00:55