TITLE: On the multiplication with constant in topological vector spaces
QUESTION [3 upvotes]: Exercise A(2) on p41 of Kelley's and Namioka's book "Linear Topological Spaces" asks to prove that if $A$ is a closed set of scalars not containing $0$ and $B$ is a closed subset of a linear topological space not containing $0$, then $AB$ is also closed.
Q1: How to prove this? (It is enough to consider only compact $A$)
Q2: Is it still true if $A$ is compact (but possibly containing $0$) and $B$ is bounded?
Q3: Any reference that I could quote regarding these facts?
Thank you.

REPLY [3 votes]: Suppose $x \in \overline{AB}$. Then there is a net $\bigl((a_{\tau},b_{\tau})\bigr)_{\tau\in T}$ in $A\times B$ such that $a_{\tau}b_{\tau} \to x$. If the net $(a_{\tau})_{\tau \in T}$ contains a bounded subnet (and it must, as we will see), it contains a convergent subnet ($\mathbb{R}$ and $\mathbb{C}$ are locally compact), so we can without loss of generality assume that $a_{\tau} \to a$, and since $A$ is closed, we have $a\in A$. Then
$$b_{\tau} = a_{\tau}^{-1}\cdot (a_{\tau}b_{\tau}) \to a^{-1}x,$$
whence by closedness of $B$ it follows that $a^{-1}x \in B$ and thus $x = a(a^{-1}x) \in AB$.
If $(a_{\tau})$ didn't contain a bounded subnet, then $a_{\tau}^{-1} \to 0$, and hence
$$b_{\tau} = a_{\tau}^{-1}\cdot(a_{\tau}b_{\tau}) \to 0\cdot x = 0,$$
contradicting the assumption that $B$ is a closed set not containing $0$.
The argument using filters instead of nets is a little less convenient.
If $A$ is compact and $B$ a bounded closed set, then $AB$ is closed, whether or not $0\in A$ or $0\in B$. If neither set contains $0$, the argument above works unmodified. Otherwise, for $x\in \overline{AB} \setminus \{0\}$ we note that by the boundedness of $B$ the net $(a_{\tau})$ cannot have $0$ as an adherent point (for then $0$ would be an adherent point of $a_{\tau}b_{\tau}$), so after extracting a subnet we have $a_{\tau} \to a \in A\setminus \{0\}$ and the rest of the argument is exactly as above. For $0 \in \overline{AB}$, we note that then $0 \in AB$ since neither $A$ nor $B$ can be empty, and we have $0 = 0\cdot b$ for every $b\in B$ in case $0\in A$, and $0 = a\cdot 0$ for every $a\in A$ in case $0\in B$.
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