Is every convergent sequence in a topological vector space bounded?

Question

It is known that in a metric space, every convergent sequence is bounded. In a topological vector space (TVS), one also has the notion of bounded sets.

Here is my question:

Is every convergent sequence in a topological vector space bounded?

---

Let $\{x_n\}$ in a TVS $X$ with $\lim x_n=0$. I think one can use the property that every neighborhood of $0$ is absorbing (and thus every finite set of $X$ is bounded) to show that $\{x_n\}$ is bounded. But I don't know how to do the general case.

Answer 1

The set of points of a convergent sequence, together with the limit of the sequence (any limit if the space isn't Hausdorff), form a (quasi)compact set in any topological space. For if we have an open covering $\mathscr{U}$ of $S := \{ x_n : n \in \mathbb{N}\} \cup \{\lambda\}$, where $x_n \to \lambda$, then there is an $U_{\lambda} \in \mathscr{U}$ with $\lambda \in U_{\lambda}$. By definition of convergence, the set $E = \{n \in \mathbb{N} : x_n \notin U_{\lambda}\}$ is finite, and for every $e\in E$, we can choose an $U_e\in \mathscr{U}$ with $x_e \in U_e$. Thus the finite family $\{U_{\lambda}\} \cup \{ U_e : e \in E\}$ covers $S$. Since $\mathscr{U}$ was an arbitrary open covering of $S$, every open covering of $S$ has a finite subcover, i.e. $S$ is (quasi)compact.

Since neighbourhoods of $0$ are absorbing, (quasi)compact sets in a topological vector space are bounded. Let $K \subset X$ be (quasi)compact. If $U$ is any neighbourhood of $0$, there is a balanced open neighbourhood $V$ of $0$ with $V \subset U$, and

$$X = \bigcup_{n \in \mathbb{N}} n\cdot V$$

since $V$ is absorbing. Thus $\{ n\cdot V : n \in \mathbb{N}\}$ is an open covering of $K$. By quasicompactness, it has a finite subcover $\{ n\cdot V : n \in F\}$. Since $V$ is balanced,

$$\bigcup_{n\in F} n \cdot V = (\max F)\cdot V,$$

and so $K \subset m\cdot V \subset m\cdot U$ for $m = \max F$. Since $U$ was arbitrary, this shows that $K$ is bounded.

And any subset of a bounded set is bounded, naturally.

Answer 2

I'm trying to give a proof by just using some basic properties of bounded sets:

• If $A$ and $B$ are bounded in $X$, then $A+B$ is also bounded;
• Every finite subset of $X$ is bounded.

Suppose $x_n\to 0$. Let $V$ be a balanced $0$-neighborhood. There exists a positive integer $N$ such that $x_n\in V$ for $n>N$. Since $S_N:=\{x_n:n=1,\cdots,N\}$ is bounded (any finite set is bounded in a TVS), we have $$ S_N\subset\lambda_1 V\subset (1+\lambda_1)V $$ for some $\lambda_1>0$ where we use the fact that $tV\subset t'V$ if $t\leq t'$ for balanced $0$-neighborhoods. Note that $\{x_n\}_{n>N}\subset V\subset (1+\lambda_1)V$. Thus $\{x_n\}_{n\geq 1}$ is bounded.

Suppose in general $x_n\to x$ for some $x\in X$. Since $V$ is a $0$-nbhd iff $V+x$ is an $x$-nbhd, one has $x_n-x\to 0$. By what we have shown above, $\{x_n-x\}$ is bounded. Since $\{-x\}$ is bounded and $$ \{x_n\}=\{x_n-x\}+\{x\}, $$ We can conclude that $\{x_n\}$ is also bounded since the Minkowski sum of two bounded sets is bounded.