Prove that $A-A = X$, where A is subset of Banach Space and dense $G_{\delta}$ set.

Question

$X$ is Banach Space and $A\subseteq X$. $A$ is dense $G_{\delta}$ set. We have to show that $A-A=X$.

Since $A$ is dense we can write $\bar{A}= X$. So, it is enough to show that $A-A= \bar{A}$. Here $A-A = \{y-z | y,z\in A\}$.

Consider $\tau$ is the induced topology by $X$. Since $X$ is normable, then $\tau$ has a bounded convex nbd of 0.

Let $V$ be a bounded, balanced, convex nbd of 0. So, $A= \cap_{r\in \mathbb{Q} } rV$.

From this I can claim that $A-A\subseteq \bar{A}$. How to show other side?

Any alternative methods also appreciated.

Answer 1

As $A$ is a dense $G_\delta$ set the complement $X\setminus A$ is of first category. Let $x_0 \in X$ and set $h:X \to X$, $h(x)=x+x_0$. Then $h$ is a homeomorphism, and with $X\setminus A$ also $X\setminus h(A)=h(X\setminus A)$ is of first category in $X$. Thus $(X\setminus A) \cup (X\setminus h(A)) \not=X$ which means $A \cap h(A) \not= \emptyset$. Choose $a \in A \cap h(A)$. Then $a=h(b)$ for some $b \in A$ which means $x_0=a-b$. Thus each $x_0 \in X$ is a difference of elements of $A$.

Answer 2

This answer speaks of subsets of a Banach space $X$. The first two exercises are results that don't hold for arbitrary topological spaces.

An exercise to get you started:

Suppose $U_n$ are all open and dense, and suppose that $U_{n+1}\subseteq U_n$. In other words, suppose that $U_n$ is a descending sequence of open dense sets.

Then $\bigcap_{n\geq1} U_n$ is non-empty.

Another exercise:

Suppose $S_n$ and $T_n$ are both descending sequences of open dense sets.

Then $(S_n\cap T_n)$ is a descending sequence of open dense sets.

Combine these results for the following:

Let $A$ be a dense $G_\delta$ set and for an arbitrary $v\in X$, let $v+A:=\{v+a\;|\;a\in A\}$.

Then $A\cap(v+A)$ is non-empty.