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I came across a lemma in some online notes where it says that, "Given a collection of elements of a basis ($\mathcal{B}$), they are also elements of Topology ($\mathcal{T}$) on $X$. How do we know this. I know we can define open sets in terms of basis. How can we go the other way?

I am using the following definition:

If $X$ is a set, a basis for a topology on $X$ is a collection $\mathcal{B}$ of subsets of $X$ (called basis elements) such that

(1)For each $x∈X$, there is at least one basis element $B$ containing $x$

(2)If $x$ belongs to the intersection of two basis elements $B_1$ and $B_2$, then there is a basis element $B_3$ containing $x$ such that $B_3⊆B_1∩B_2$.

If $\mathcal{B}$ satisfies these two conditions, then we define the topology $\mathcal{T}$ generated by $\mathcal{B}$ as follows: A subset $U$ of $X$ is said to be open in $X$ if for each $x∈U$, there is a basis element B∈$\mathcal{B}$ such that $x∈B$ and $B⊆U$.

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    $\begingroup$ By definition, a basis of the topology consists of open sets. $\endgroup$ Oct 4, 2013 at 14:57
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    $\begingroup$ @JamesBond, $\mathcal{B}$ is some special collection of open sets. $\mathcal{B}$ is not open in itself nor is it closet since it's a collection of sets and not a subset of $X$ ! However, the union of all elements in the collection $\mathcal{B}$ is open (this union is in fact $X$) as union of open sets is open. $\endgroup$
    – M.G
    Oct 4, 2013 at 15:13
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    $\begingroup$ @JamesBond You are correct : any kind of union of elements of $\mathcal{B}$ will only produce open sets. Consider the the real line with the standard topology. You can check that a basis for this topology is the collection of all open intervals. However, $(0,1)\cup(2,3)$ is not an open interval so this set is not an element of the basis. However, it's an element of the topology (as a union of two elements of the basis). $\endgroup$
    – M.G
    Oct 4, 2013 at 15:26
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    $\begingroup$ If $B \in \mathcal{B}$, then for all $x \in B$, we have $x \in B$ and $B \subseteq B$, so $B$ is open by the definition of $\mathcal{T}$. $\endgroup$ Oct 4, 2013 at 15:30
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    $\begingroup$ @ChristianBlatter Point (2) guarantees that every intersection of basis sets is a union of basis sets. Thus it's indeed a basis. (I don't like the definition, but it's valid.) $\endgroup$ Oct 4, 2013 at 15:40

2 Answers 2

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The way you define the open sets in your edited question, it follows that each $B\in\mathcal B$ is open, since for $x\in B$ we have $x\in B$ and $B\subseteq B$.

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A basis $\mathcal{B}$ for a topological space X with topology $\mathcal{T}$ is a collection of open sets in $\mathcal{T}$ such that every open set in $\mathcal{T}$ can be written as a union of elements of $\mathcal{B}$.

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  • $\begingroup$ I am looking at a different definition of basis. It is the one that Munkres uses in his book $\endgroup$
    – James Bond
    Oct 4, 2013 at 15:06
  • $\begingroup$ I'll have a look. The definition may be different but the concepts behind are the same. $\endgroup$
    – M.G
    Oct 4, 2013 at 15:07
  • $\begingroup$ Have a look to Lemma 13.1 of Munkres' and to the paragraph that follows : "The lemma states that every open set $U$ in $X$ can be expressed as a union of basis elements". $\endgroup$
    – M.G
    Oct 4, 2013 at 15:19
  • $\begingroup$ @JamesBond: If you are using a different definition, could you please edit the question to include it? $\endgroup$ Oct 4, 2013 at 15:22
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    $\begingroup$ @JamesBond The definition you gave is equivalent to the one given by the answer above. $\endgroup$ Oct 4, 2013 at 16:20

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