Why basis of topological space is union of local basis of topological space?

Question

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.

Answer 1

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 $(X,O)$. $\blacksquare$

Answer 2

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.