# Kelley & Namioka's definition of topology of uniform convergence on a subset

In Kelley & Namioka's Linear Topological Spaces, they begin section 8 on Function Spaces with a definition of the topology of uniform convergence. I've reproduced the begining of the first paragraph below:

Let $S$ be any set, and let $E$ be a linear topological space. The set $F(S,E)$ of all functions on $S$ to $E$, with addition and scalar multiplication defined pointwise, is a linear space. For each subset $A$ of $S$, and for each neighborhood [Kelley's neighborhoods needn't be open] $U$ of $0$ in $E$, let $N(A,U)$ be the family of all members $f$ of $F(S,E)$ with the property that $f[A]\subset U$. Observe that $N(A,U)$ is circled if $U$ is circled. A subset $G$ of $F(S,E)$ is open relative to the topology of uniform convergence on $A$ if and only if for each $f$ in $G$ there is a neighborhood $U$ of $0$ in $E$ such that $f+N(A,U)\subset G$. It is easy to verify that this definition gives a topology for $F(S,E)$ such that the family of sets of the form $f+N(A,U)$ is a base for the neighborhood system of $f$.

I have no problem with this "easy" verification except for one point. Why is $f+N(A,U)$ a neighborhood of $f$ in the defined topology? To make the question even simpler, let's take $f=0$. Why is $N(A,U)$ a neighborhood of $0$? What possible open set $G$ containing $0$ could be contained in $N(A,U)$? And why is it easy to see?

I have convinced myself that, in general, $N(A,U)$ itself will not be open. A simple counterexample, which illustrates the problem I run into with showing any $G$ is open, is as follows. Let $E=R$ (the reals) and let $S=A=U=(-1,1)$. Then $E$ is a TVS and $U$ is a neighborhood of $0$ in $E$. Let $f$ be the identity map taking $A$ to $U$. We then have $f\in N(A,U)$. For $N(A,U)$ to be open, it must contain $f+N(A,V)$ for some neighborhood $V$ of $0$ in $E$. Given any such $V$, there must be an $\epsilon\in(0,1)$ such that $(-\epsilon,\epsilon)\subset V$. Let $g=\epsilon f$. Then $g\in N(A,V)$, but $(f+g)(1-\epsilon/2)=1-\epsilon/2+\epsilon-\epsilon^2/2=1+\epsilon/2(1-\epsilon)>1$, so $f+g\notin N(A,U)$. The same sort of thing seems to crop up when trying to show that there is any open $G$ containing 0 and contained in $N(A,U)$.

To use the example you gave, the interior of $N(A,U)$ consists of all functions $f$ such that $f[(-1,1)]\subseteq (-1,1)$ and such that there exists $\epsilon>0$ such that $f+N(A,(-\epsilon,\epsilon))\subseteq N(A,(-1,1))$. These are just the functions such that $\inf f[A]>-1$ and $\sup f[A]<1$.
For the general case, take a neighborhood $W$ of $0$ such that $W+W\subseteq U$. Then $N(A,W)$ contains $0$ and is a subset of the interior of $N(A,U)$. For $f\in N(A,W)$ implies that for all $g\in N(A,W)$ one has $(f+g)[A]\subseteq W+W\subseteq U$ and therefore $f+N(A,W)\subseteq N(A,U)$.
It follows that $N(A,U)$ has always a nonempty interior.
• Michael, could you please tell me in your reply, which set $G$ are you testing for openness with the definition of the topology given by K&N? It can't be the interior of $N(A,U)$ since that would presume it exists. Oct 17 '16 at 3:35
• I show that $N(A,U)$ has interior points and that $0$ is one of them. This shows that $N(A,U)$ is a neighborhood of $0$, without explicitly giving a set $G$. But a variant of the argument shows that you can take $G$ to be the union of all sets of the form $N(A,V)$ in which $V$ is a neighborhood of $0$ such that $V+W\subseteq U$ for some neighborhood $W$ of $0$. Oct 17 '16 at 3:50
• That doesn't seem to work. To expand what your proposed proof would look like: Let $f\in G$. Then for some neighborhood $V$ of $0$ in $E$, $f\in N(A,V)$. Associated with $V$ is another neighborhood $W$ of $0$ such that $V+W\subseteq U$. We must now show that $f+N(A,W)\subseteq G$. However, all we can show is that $f+N(A,W)\subseteq N(A,U)$. Oct 17 '16 at 16:50
• Actually, maybe this proof can be made to work! Instead of using $f+N(A,W)$, let $T$ be a neighborhood of $0$ such that $T+T\subseteq W$. Then $f+N(A,T)\subseteq G$ since if $g\in N(A,T)$, we have $f+g\in N(A,V+T)\subseteq G$ since $(V+T)+T\subseteq V+W\subseteq U$. Thanks, Michael!!! Oct 17 '16 at 17:02