# Proof Check for General Topology of Topological Groups

This is not for homework, I'm working through Munkre's to prepare for a class next semester and I did the following exercise (17 Section 30). Paraphrased:

Show that a first countable, separable topological group is second countable. [Hint: Let $\{B_n\}$ be a countable basis at $e$. If $D$ is a countable dense subset of $G$, then show that $dB_n$ where $d\in D$ is a basis for $G$.

To show that $\{dB_n \:|\: d \in D, n \in \mathbb{N}\}$ is a basis, I need to show that it covers $G$ and that any intersection of basis elements contains a basis element.

Here's my attempt.

Let $G$ be a first-countable, separable topological group, and choose a countable neighborhood basis of $e$, denoted $\mathcal{B} = \{B_n\}_{n\in \mathbb{N}}$. Let $x \in G$ and consider the left multiplication map $l_x$ which is a homeomorphism, thus $l_x(B_n)$ is open for all $B_n \in \mathcal{B}$. Further, consider the inversion map $i: G \to G$ (which is again a homeomorphism since $G$ is a topological group). As $B_n$ is open and $i$ is continuous, choose an open set $U \subset G$ about $e$ such that $i(U) \subset B_n$. Let $V = U \cap B_n$ (which is nonempty since $e \in U$ and $e \in B_n$), thus $i(V) \subset i(U) \subset B_n$. Since $V$ and $i(V)$ are open, contained in $B_n$, and contain the identity, we have that $W= V \cap i(V)$ is a nonempty open subset of $B_n$ containing $e$. Consider $l_x(W) = xW$ which is open in $G$ and contains $x$ since $x = x \cdot e$ and $e\in W$. As $xW$ is open and $D$ is dense, $W$ contains some $d \in D$, thus $d= xw$ for some $w \in W$. Since $W = V \cap i(V)$, we have that $w^{-1} \in W$, and $x = dw^{-1} \in dW \subset dB_n$. As $x$ is arbitrary, we have that $dB_n$ covers $G$.

Let $d,d' \in D$ and consider the set $dB_n \cap d'B_m$. If the intersection is empty, then we're done, otherwise let $x \in dB_n \cap d'B_m$. As $dB_n \cap d'B_m$ is open, and the map $l_x$ is continuous, there exists a neighborhood $U\subset G$ about $e$ such that $l_x(U) \subset dB_n \cap d'B_m$. Let $V = B_n\cap B_m \cap U$ which is open, as it is a finite intersection of open sets, and nonempty, since $e \in B_n \cap B_m \cap U$, thus, by the definition of a basis, $B_n \cap B_m \cap U$ contains a basis element $B \in \mathcal{B}$. Finally, consider the nonempty open set $W= B \cap i(B)$ and let $B' \subset W$ be a basis element, which, by transitivity is a subset of $B_n \cap B_m \cap U$, and contains $e$. Since $l_x$ is a homeomorphism, $l_x(B')$ is open, thus contains some $d'' \in D$, i.e. there is some $b' \in B'$ such that $xb' = d''$, i.e. $x = d''b'^{-1} \in d''B'$, and $d''B' \in \{dB_n \:|\: d \in D \text{ and } B_n \in \mathcal{B}\}$, and this set forms a countable basis for $G$.

It's quite terse, so that's question number 1, can this be simplified so as to ease notation and density?

My main concern however is about the construction of the set $W$. First, I use the continuity of $i$ to define a neighborhood $U$ of $e$ such that $i(U) \subset B_n$. Is this step allowed? And does $B_n$ necessarily contain $e$?

Finally, are there are any other errors I failed spot?

• You have to show that for each $x \in G$ and each open neighborhood $U$ of $x$ there exist $n$ and $d \in D$ such that $x \in dB_n \subset U$. Try to do this. Note that it suffices to consider $U = xB_m$ for some $m$. Jul 24, 2018 at 13:17
• Okay this simplifies the two part argument above. I'll correct it and answer my own question tonight perhaps.
– Nico
Jul 24, 2018 at 17:17

We have to show that for each $x \in G$ and each open neighborhood $U$ of $x$ there exist $n$ and $d \in D$ such that $x \in dB_n \subset U$.
We have $x \in U$, therefore $e \in x^{-1}U$. Since the function $(x,y) \mapsto x^{-1}y$ is continuous, we find $B_n$ such that $B_n^{-1}B_n \subset x^{-1}U$. Choose $d \in xB_n^{-1} \cap D$. Hence $x^{-1}d \in B_n^{-1}$, $d^{-1}x \in B_n$ and $x \in dB_n \subset xB_n^{-1}B_n \subset xx^{-1}U = U$.
• I may be missing something obvious. I don't see why $dB_n\subset xB_n^{-1}B_n.$ Jul 27, 2018 at 4:32
• @DanielWainfleet We chose $d \in xB_n^{-1}$, therefore $dB_n \subset xB_n ^{-1}B_n$. But now I see that it is unnecessary to introduce $B_m$. I shall edit my proof. Jul 27, 2018 at 8:31
• Thank you. I think it $was$ obvious but not to me. Jul 29, 2018 at 6:32