Suppose that we have a normed vector space $(X,\|\cdot\|)$. We endow $X$ with a (locally convex) topology $\tau$ such that any $\tau$-convergent is norm bounded. Suppose that there exists a countably infinite subset $A=\{x_{1},x_{2},\ldots\}\subset X$ that satisfies the following:
- $\|x_{i}\|\to\infty$,
- $0\notin A$,
- $0\in\overline{A}^{\tau}$.
I want to prove that $X$ cannot be $\tau$-first-countable under the above assumptions.
I tried to argue by contradiction. If $X$ is $\tau$-first-countable, then there exists a sequence $(a_{k})$ that $\tau$-converges to $0$. But then $(a_{k})$ must have finitely many different coordinates, otherwise it is unbounded by the assumption on $A$ and bounded by the assumption on $\tau$-convergent sequences in $X$, which is absurd.
But I don't know if my arguments are right. Furthermore, I don't know how to finish.
Maybe this follows from a more general (topology) result. Any help would be greatly apreciated!
Best Answer
Suppose that $X$ is first countable. As you said, then there is a sequence $(a_k)\subset A$ with $a_k\to0$. Since $(a_k)$ is convergent, it is norm-bounded. Obviously, we have $a_k=x_{n_k}$ for all $k$, where $n_k\in\mathbb{N}$ are not necessarily distinct, nor in increasing order, i.e. $(a_k)$ is not necessarily a subsequence of $(x_n)$. If there where infinitely many different $a_k$, then we would have that $(a_k)$ contained an unbounded subsequence in norm (this is easily verified) which is impossible. So, there are finitely many different elements $a_k$. So, the sequence stabilizes after a certain point, i.e. there is $k_0\in\mathbb{N}$ such that $a_k=a_{k_0}$ for all $k\geq k_0$. BUt since $a_k\to0$, we have that $a_k=0$ for $k\geq k_0$. This implies that $0\in A$, a contradiction.
Since our assumption that $X$ is first countable led to a contradiction, $X$ is not first countable (and you have actually already finished!)
Your reasoning is correct, after the edit on $A$.