Exercise 71.4

Proof. Let $p\in X$ be the common point of the circles. Suppose $X$ has a countable basis $\{U_i\}$ at $p$; we can assume without loss of generality that $U_i⊋U_j$ if $i<j$. For each $U_i$, we know that $V_{\mathit{ij}}=U_i\cap S_j$ is open for any $i,j$ by the coherence condition, and is nonempty since $p\in V_{\mathit{ij}}$.

Now take $V=\bigcup <!--\; FUNCTION\; APPLICATION\; -->V_{\mathit{ii}}$; we claim it is a neighborhood of $p$ that does not contain any $U_i$. It is open by coherence since $V\cap S_i=U_i$ is open for all $i$, and contains $p$ by construction above. Now suppose $U_i\subseteq V$. Then, this implies $U_i\cap S_j\subseteq V\cap S_j=U_j$ for all $j$, and in particular when $i<j$, which contradicts that $U_i⊋U_j$ if $i<j$. □