Problem: To show that, in a topological vector space, for a given neighborhood of zero $W$, there exist two neighborhoods of zero, $V_1$, $V_2$, whose sum is contained in the first neighborhood, $V_1+V_2\subset W$.

Background: I am reviewing some foundations of functional analysis and was looking at Rudin's book. Near the beginning he states his theorem 1.10. Its point is to show that if we start with two disjoint sets in a topological vector space $X$: a compact set K and a closed set C, then there are disjoint open sets containing them. However, I got stuck in a small detail of the proof.

He shows first that if $W$ is a neighborhood of $0$ (the zero vector), then there is a neighborhood $U$ of $0$ (the zero vector) which is symmetric (in the sense that $U = - U$) and which satisfies $U + U \subset W$. He says: we note that $0+ 0 = 0$, that addition is continuous, and that $0$ therefore has neighborhoods $V_1$, $V_2$ such that $V_1 + V_2 \subset W$. etc...

This last statement I do not get. It is clear that $W$ as an open set contains other open sets (for example itself). I can construct, using the continuity of the multiplication by scalars, many other sets properly contained in $W$, contracting and, if needed, taking intersections with $W$. But it is not clear to me how to construct (or show that exist) sets with addition still in $W$.

If the vector space is finite or normed or convex this is not hard to do. In the normed case I simply take the balls centered at the origin that are half the size of the least norm of the elements of $W$, but he is stating this in the general topological vector space case. I have not found the generalization of my procedure to the general setting.

I also tried looking at all possible pairs of $0-$neighborhoods contained in $W$. If every single pair of open sets adds up to a set with nonempty intersection with $W$ then every pair of sets give a pair of points $x$, $y$ with sum $x+y$ not in $W$. I would like to show this eliminates all points but the origin (but I I am not sure I can do this). Then either we reach a contradiction, or are forced to consider $0+0=0$ as counterexample which works in the case of the discrete topology in $X$.

Since the statement is so brief, I am afraid I am missing something. Thank you for your help.


The function $f(x,y) = x+y$ is continuous, and $f(0,0) = 0$. Let $W$ be an open set containing $0$, then by the definition of continuity, there exists open sets $V_1,V_2$ such that $0 \in V_1$, $0 \in V_2$ such that $f(x,y) \in W$ for all $x \in V_1$, and $y \in V_2$. In other words, $V_1+V_2 \subset W$.

  • $\begingroup$ Just a small clarification. The set $+^{-1}(W)$ is open in $X\times X$ and the topology of that space is built from products of open sets, so it must contain sets of the form $V_1\times V_2$ from which we extract $V_1$, $V_2$ by projection, but we still need to force them to live in $W$. I think that taking the subset $(W\times W) \cap +^{-1}(W)$ does construct a suitable open set in $X\times X$ from which we can find $V_1\times V_2$. $\endgroup$ – F. Solis Jan 22 '13 at 18:02
  • $\begingroup$ I'm not sure what you mean by the 'force to live in $W$' part. The set $f^{-1} W$ is open by definition, and if $(x,y) \in f^{-1} W$, then $x+y \in W$, of course. So, the only issue is to show that $f^{-1} W$ contains a neighborhood of $(0,0)$ of the form $V_1 \times V_2$. But there is no need, a priori, to take the intersection with $W^2$. $\endgroup$ – copper.hat Jan 22 '13 at 18:08
  • $\begingroup$ Sorry, I didn't finish my line of thought: The sets $U_1 \times U_2$ (where $U_i \in \tau_V$) form a basis for the product topology on $V\times V$, hence $f^{-1}W$ can be written as the union of base elements, at least one of which contains $(0,0)$. Let $V_1\times V_2$ be the latter element. $\endgroup$ – copper.hat Jan 22 '13 at 18:30
  • $\begingroup$ If $x+y\in W$, is not always the case that $x \in W$ and $y\in W$, which is part of the statement. $\endgroup$ – F. Solis Jan 22 '13 at 19:41
  • $\begingroup$ If $x+y\in W$, is not always the case that $x \in W$ and $y\in W$, which is part of the statement. As an example, $W=(-a,a)$ in $X$ the real line. $+^{-1}(W)$ is an infinite stripe in two dimensions $-a<x+y<a$. Take $a=1$ and see that $V_1=(-2.2,-1.8)$ and $V_2=(1.8,2.2)$ satisfy $V_1+V_2=W$ but are not in $W$. Therefore, a subset of $+^{-1}(W)$ has to first be limited before extracting $V_1$ and $V_2$. $\endgroup$ – F. Solis Jan 22 '13 at 19:50

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