Balanced Set in Topological Vector Space is Connected

Theorem

Let $\GF \in \set {\R, \C}$.

Let $X$ be a topological vector space over $\GF$.

Let $B \subseteq X$ be a balanced set.

Then $B$ is connected.

Proof

We first show that $B$ is path connected.

Let $x \in B$.

Define $p : \closedint 0 1 \to X$ by:

$\map p t = t x$

We clearly have:

$\map p 0 = 0$

and:

$\map p 1 = x$

Since $B$ is balanced, we have:

$t x \in B$

for each $t \in \closedint 0 1$.

It remains to show that $p$ is continuous.

Define $m : \closedint 0 1 \times X$ by:

$\map m {\lambda, x} = \lambda x$

for each $\tuple {\lambda, x} \in \closedint 0 1 \times X$.

From Restriction of Continuous Mapping is Continuous, we have that $m$ is continuous.

We can then see that $p$ is the $x$-horizontal section of $m$.

From Horizontal Section of Continuous Function is Continuous, we have that $p$ is continuous.

So every $x \in B$ connects to $0 \in B$.

So from Path-Connected iff Path-Connected to Point, we have that $B$ is path-connected.

From Path-Connected Space is Connected, $B$ is connected.

$\blacksquare$