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Let $(X,O)$ be topological space which is defined by local basis(fundamental neighborhood system)$B_x$ at $x∈X$. Then, why $∪_{x∈X}B_x$ is basis of $X$, in other words, every open set of $(X,O)$ is written by union of $∪_{x∈X}B_x$?

I think this holds in general topological space, but I have never seen this statement in any my text books or in websites. Could you tell me the proof of this or reference?

Thank you in advance.

Martin Sleziak
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Poitou-Tate
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2 Answers2

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Let $(X,O)$ be a topological space.

  • Given $x \in X$, a local basis at $x$ is a family $\mathcal B_x$ of elements of $O$ with the following two properties:
    • $x$ is in every element of $\mathcal B_x$; and
    • for every $U \in O$ with $x \in U$ there exists $B \in \mathcal B_x$ such that $B \subseteq U$.
  • A basis of $(X,O)$ is a family $\mathcal B$ of elements of $O$ with the following property: For every $U \in O$ and every $x \in U$ there must be exists $B \in \mathcal B$ such that $x \in B \subseteq U$. (Equivalently, $\mathcal B$ is a basis of $(X,O)$ if every element of $O$ is a union of elements of $\mathcal B$.)

Claim: If for any $x \in X$ we have a local basis at $x$, say $\mathcal B_x$, then $\mathcal B := \bigcup_{x \in X} \mathcal B_x$ is a basis of $(X,O)$.

Proof: Take $U \in O$ and $x \in U$. By the second property in the definition of local basis we know that there exists $B \in \mathcal B_x$ such that $B \subseteq U$. Now, by the first property in the definition of local basis we know that $x \in B$. Therefore $x \in B \subseteq U$. Since $B \in \mathcal B_x$ says that $B \in \mathcal B$, we have proved the following: For every $U \in O$ and every $x \in U$ there must be exists $B \in \mathcal B$ such that $x \in B \subseteq U$. Hence $\mathcal B$ is a basis of $(X,O)$. $\blacksquare$

azif00
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For a set $\mathcal{B}$ to be a basis for a topological space $\left( X, O \right)$, we require the following:

  1. The union of all elements of $\mathcal{B}$ should be all of $X$.
  2. Given elements $B_1, B_2 \in \mathcal{B}$ with $x \in B_1 \cap B_2$, we should be able to find $B \in \mathcal{B}$ such that $x \in B \subseteq B_1 \cap B_2$.

It is clear as to why the union of all elements of $\bigcup\limits_{x \in X} B_x$ is all of $X$. Particularly, it is true because each $x \in X$ is contained in every element of $B_x$ and hence together they all cover $X$.

For the second part, we notice that a local base is made of open sets. Therefore, both $B_1$ and $B_2$ are open, and each contains $x$. Using he properties of base, we can find $B_1', B_2' \in B_x$ such that $x \in B_1' \subseteq B_1$ and $x \in B_2' \subseteq B_2$. Consequently, since now both $B_1', B_2' \in B_x$, there is an element $B \in B_x$ such that $x \in B_1' \cap B_2' \subseteq B_1 \cap B_2$.

This tells us that the union of all local bases is indeed a base for the topology.

Aniruddha Deshmukh
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  • There's a small typo when x\subset should be x\in B, in the last expression. I still don't see why this " Using he properties of base, we can find B′1,B′2∈Bx such that x∈B′1⊆B1 and x∈B′2⊆B2. " but i´m on it – Weronika Mar 17 '22 at 05:24
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    @VerónicaR.M.: That is simply because $B_1$ is an open set containing $x$ and $B_x$ is a local base at $x$. Recall the definition of a local base and apply it. – Aniruddha Deshmukh Mar 17 '22 at 05:34
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    Your answer only proves that $\mathcal B$ satisfies the conditions for being a basis for some topology in $X$, it does not prove that it is a basis for $(X,O)$. See my answer. – azif00 Mar 17 '22 at 05:42
  • @azif00: You are correct! – Aniruddha Deshmukh Mar 17 '22 at 05:43