# Lemma 13.2 from Munkres

1. Aug 1, 2011

### camilus

Lemma 13.2: Let X be a topological space. Suppose that C is a collection of open sets of X such that for each open set U of X and each x in U, there is an element c of C such that $x\in c\subset U$. Then C is a basis for the topology of X.

Proof: The first paragraph is trivial, it just shows that the conditions of basis are satisfied.

The second paragraph attempts to show that $\tau'$, the topology generated by C, is the same as the topology $\tau$ on X.

Can someone elaborate on the second paragraph please!?!

2. Aug 1, 2011

### micromass

Staff Emeritus
What don't you understand about the second paragraph??

You just let $\mathcal{T}^\prime$ be the topology generated by the basis $\mathcal{C}$. So by definition, a set G belongs to $\mathcal{T}^\prime$ if for every x in G, there is a set C in $\mathcal{C}$ such that

$$x\in C\subseteq G$$

So, to prove that $\mathcal{T}\subseteq\mathcal{T}^\prime$. Take G in $\mathcal{T}$. By hypothesis, there exists for every for every x in G, a set C in $\mathcal{C}$ such that

$$x\in C\subseteq G$$

So, by definition almost, we have that $\mathcal{T}\subseteq \mathcal{T}^\prime$.

The other inclusion is less obvious. So to prove that $\mathcal{T}^\prime\subseteq\mathcal{T}$ we take an element $G\in \mathcal{T}^\prime$. By hypothesis, there is for every x in G, a set C in $\mathcal{C}$ such that

$$x\in C\subseteq G$$

This implies that $G=\bigcup{C}$ is the union of all these C's. Since all the C's are in $\mathcal{T}$, it implies that G is also in $\mathcal{T}$ as union of open sets...

IS that more clear?