# If $X$ is complete and $T$ is continuous on $X$, is $T(X)$ complete?

I have this problem: If $X$ is finite-dimensional normed linear space and $Y$ is a normed linear space, then every linear operator $T:X\rightarrow Y$ is continuous and open.

I have already shown that $T$ is continuous. Of course $X$ is complete. However, I cannot use the Open-Mapping Theorem since $Y$ is not necessarily complete.

My idea now is if I use $T(X)$ instead of $Y$, I can use the open-mapping theorem because $T$ is now onto $T(X)$. But, is $T(X)$ complete? Or maybe $T(X)$ finite-dimensional as well, just like $X$?

But $T$ doesn't have to be open. Take $T\colon\mathbb{R}\longrightarrow\mathbb{R}^2$ defined by $T(x)=(x,0)$, for instance.
• Maybe because $T$ here is not surjective. Apr 11, 2018 at 9:00
• I don't understand your comment. My map $T$ isn't open because it doesn't map open sets onto open sets. Why did you mention surjectivity? Apr 11, 2018 at 9:11
Since $T$ is linear thus $T(X)$ is finte dimensional but every finite dimensional linear space is complete.
• Since every finite dimensional space over the field $\mathbb{R}$ or $\mathbb{C}$ is locally compact. But every locally compact metric space is complete.