Question about summable sequences

Question

Let $(E, ||\cdot||)$ be a normed vector space and $(x_\alpha)_{\alpha \in A}\subset E$ be a summable family. Then ,the set $B = \{\alpha \in A | x_\alpha \ne 0\}$ is at most countable.

I am stuck with the proof. What I've tried:

Because $(x_\alpha)_\alpha$ is summable, it is Cauchy, so for every $\epsilon > 0$ there exists $F_\epsilon \subset A$ a finite set such that for any $J \subset A$ finite such that $F_\epsilon \cap J = \emptyset$ we have $||\sum_{\alpha \in J} x_\alpha|| < \epsilon$.

Now, let's take $\epsilon = \frac{1}{n}$. Then there exists $F_n \subset A$ finite such that for any $J \subset A$ with $J \cap F_n = \emptyset$ we have $||\sum_{\alpha \in J} x_\alpha|| < \epsilon$.

We can observe that $B = \cup_{n \in \mathbb{N}}B_n$ where $B_n = \{\alpha \in A | ||x_\alpha|| > \frac{1}{n}\}$. Now suppose that $B_n$ is infinite. Then we can take $C=\{\alpha_1, \ldots, \alpha_k\} \subset B_n \setminus F_n$. So, $C \cap F_n = \emptyset$ and $C$ is finite implies $||\sum_{\alpha \in C}x_\alpha|| < \frac{1}{n}$. But $\sum_{\alpha \in C} ||x_\alpha|| > \frac{k}{n}$.

How I can get here a contradiction? (Because if I get a contradiction, then B is countable union of at most countable sets, so it is at most countable and the proof is done).

Answer 1

You were close to completing the proof with the sets $F_n$ you introduced.

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To prove that $B$ is at most countable, you can simply show that it is contained in the union of the $F_n$ (which is obviously at most countable since each $F_n$ is finite), or equivalently, that the complement in $A$ of the union of the $F_n$ is contained in that of $B$.

So suppose that $\alpha\in A\setminus\bigcup_{n\in\mathbb N}F_n$. The set $J=\{\alpha\}$ satisfies $J\cap F_n=\emptyset$ for every $n$, so $$\|x_\alpha\|=\left\|\sum_{\beta\in J}x_\beta\right\|<\frac 1n$$ for every $n$. This tells you that $\|x_\alpha\|=0$ and then that $x_\alpha=0$ and then that $\alpha\in A\setminus B$.

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In a very similar fashion, you can prove the following result.

Let $G$ be an abelian Hausdorff topological group and let $\mathfrak B$ be a basis of the filter of neighbourhoods of $0$ in $G$. If $(x_\alpha)_{\alpha\in A}$ is a family of elements of $G$ that is summable in $G$, then the cardinality of the set $\{\alpha\in A~|~x_\alpha\neq 0\}$ is at most that of $\mathfrak B$.