Is every convergent sequence in a topological vector space bounded? 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. 

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. 
