# How to prove range of the following operator is closed?

If $X$ is an infinite dimensional hilbert space and $X_n$ be its $n$ dimensional subspace. Let $P$ denotes the orthogonal projection on $X_n$ from $X$. Also $T$ is a bounded linear operator from $X$ to hilbert space $Y$.

Now How to prove that range of this map $TP:X \to Y$ is closed.

I know the following results

1. For an orthogonal projection range$(P)$ is closed.

2. the range of $T$ is closed in $Y$ if $T$ is open.

But how to prove this i don't know.

• A finite dimensional subspace is closed. – copper.hat Aug 10 '16 at 14:54

The range of $TP$ is a finite dimensional submanifold, so it is closed.
• can you please explain how? I have Range $P$ is finite dimensional but then after that? – kapil Aug 10 '16 at 15:11
• yes i know this result. But my doubt is which property of $T$ we are using here? Thanks in advance – kapil Aug 10 '16 at 15:15
• If $S$ is a finite dimensional subspace, then so is $TS$. – copper.hat Aug 10 '16 at 15:47