Why is a subrepresentation a *closed* invariant subspace?

Question

If $(\rho,V)$ is a topological representation of a topological group $G$, we usually say that a subrepresentation consists of a closed subspace $U\subset V$ which is $G$-invariant. I don't understand why it has to be closed. If \begin{align*} G\times V &\to V\\ (g,v)&\mapsto \rho(g)v \end{align*} is continuous, then so is the restriction $G\times U\to U$ even if $U$ is not closed. In other words, if we omit the need for $U$ to be closed, $(\rho,U)$ is still a topological representation. So why we impose this condition?

Answer 1

The motivation is that closed subspaces allow nice decompositions in infinite dimensions (obviously in finite dimensions all subspaces are closed). If you have a Hilbert space $H$ and a closed subspace $V \subset H$ then $H = V \oplus V^\perp$. If $V$ were not closed this wouldn't be true. For example, suppose $V$ were a dense subspace of $H$ not equal to $H$. Then $V^\perp=0$ and so clearly $V \oplus V^\perp \neq H$.