# Alternate definition for boundedness in a TVS

Let $X$ be a topological vector space over $\mathbb R$ or $\mathbb C$. A subset $B\subset X$ is defined to be bounded if

for any open neighborhood $N$ of $0$ there is a number $\lambda>0$ such that $B\subset \mu N$ for any $\mu>\lambda$.

I was wondering whether this notion of boundedness is equivalent to saying that for any open neighborhood $N$ of $0$ there is a $\mu>0$ such that $B\subset \mu N.$

By definition $\mu N:=\{\mu\cdot x\colon x\in N\}$.

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You seem to ignore $N$ in the definition – Dennis Gulko Oct 2 '12 at 12:24

The usual definition implies the second one.

To see the other, we can restrict ourselves to balanced neighborhood, that is, to sets $V$ such that $\alpha V\subset V$ for $|\alpha |<1$. Indeed, each open non-empty set contains a balanced neighborhood, by continuity of the map $(\lambda,x)\mapsto \lambda\cdot x$. So we can find $r$ and $U\subset V$ open such that $\alpha U\subset V$ if $|\alpha|<r$. Then $W:=\bigcup_{|\alpha|<r}\alpha U$ does the job: it's open as a countable union of such sets and balanced by definition.

Take a neighborhood $N$ of $0$, then $V\subset N$ a balanced neighborhood of $0$. We can find $\lambda>0$ such that $B\subset \lambda V$. Take $\mu>\lambda$. Then $0<\frac{\lambda}{\mu}<1$ and $\frac{\lambda}{\mu}V\subset V$. Hence $$B\subset \lambda V\subset \mu V\subset \mu N$$ and we are done.

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In a locally convex space the two definitions are easily equivalent because there exists a convex neighborhood base and any neighborhood contains $0$.

However in general two two definitions are equivalent since you always have a balanced neighborhood base.

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