Hatcher Universal Covering Space Construction - Basis

Question

Below is an excerpt from Hatcher's Algebraic Topology. He is constructing a universal cover for a path-connected, locally path-connected, and semilocally simply-connected space $X$:

I don't understand why it follows that $\mathcal{U}$ is a basis for $X$. Is this a general fact that if a collection of open sets claimed to be a basis has the property that every basis element contains another basis element, then the collection is actually a basis? This doesn't sound right.

Answer 1

A basis for a topology $\mathcal{T}$ on a space $X$ is a subset $\mathcal{B} \subset \mathcal{T}$ such that for each $U \in \mathcal{T}$ and each $x \in U$ there exists $B \in \mathcal{B}$ such that $x \in B \subset U$.

A space $X$ is defined to be locally path connected if it has a basis consisting of path connected open sets (in other word, if the set $\mathcal{P}$ of path connected open sets forms a basis for $X$).

Hatcher shows that in a locally path connected semilocally simply-connected space $X$ the subset $\mathcal{U} \subset \mathcal{P}$ of all $U \in \mathcal{P}$ such that $\pi_1(U) \to \pi_1(X)$ is trivial also forms a basis for $X$. Note that this part does not use that $X$ is path connected.

Answer 2

Let me iron all of this out.

Let $U$ be an open set of $X$ and $x \in U$. Since $X$ is locally path connected, $X$ has a basis $\mathcal{B}$ of path connected sets. Hence, we can find a $B_1 \in \mathcal{B}$ such that $x \in B_1 \subset U$. Since $X$ is semilocally simply connected, there is a an open set $V$ containing $x$ such that the induced inclusion $\pi_1(V,x) \rightarrow \pi_1(X,x)$ is trivial. Hence, we can find a $B_2 \in \mathcal{B}$ such that $x \in B_2 \subset V$ and furthermore we can find a $B_3 \in \mathcal{B}$ such that $x \in B_3 \subset B_1 \cap B_2 \subset V$. Then the composition of induced inclusions $\pi_1(B_3,x) \rightarrow \pi_1(V,x) \rightarrow \pi_1(X,x)$ is trivial and $B_3 \subset U$. Thus, what we have shown is that the subset of $\mathcal{B}$ consisting of elements $W$ such that the induced inclusion $\pi_1(W) \rightarrow \pi_1(X)$ is trivial is also a basis for $X$.

(Remember that since $W$ is path connected and the induced inclusion $\pi_1(W,x) \rightarrow \pi_1(X,x)$ is trivial for the choice of basepoint $x$ above, the induced inclusion is trivial for all other basepoints in $W$. Hence, it is acceptable to omit the basepoint and write that the induced inclusion $\pi_1(W) \rightarrow \pi_1(X)$ is trivial.)