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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 $X$?

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2 Answers 2

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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.

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  • $\begingroup$ Maybe because $T$ here is not surjective. $\endgroup$
    – admath
    Apr 11, 2018 at 9:00
  • $\begingroup$ 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? $\endgroup$
    – José Carlos Santos
    Apr 11, 2018 at 9:11
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Since $T$ is linear thus $T(X)$ is finte dimensional but every finite dimensional linear space is complete.

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  • $\begingroup$ Thanks for this. But why is it true? $\endgroup$
    – admath
    Apr 11, 2018 at 8:56
  • $\begingroup$ Since every finite dimensional space over the field $\mathbb{R}$ or $\mathbb{C}$ is locally compact. But every locally compact metric space is complete. $\endgroup$
    – user235708
    Apr 11, 2018 at 8:58

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