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Let $T\colon C[0,1]\to C[0,1]$ be a bijective operator. My (maybe silly) question is if $T$ then has a dense image.

I think I can show this in general: Let $f\colon X\to Y$ be bijective. Consider any $y\in Y$. Then there is exactly one $x\in X$ with $y=f(x)$. So of course $y\in\mbox{rn}(f)$. And so chosse the sequence $(x_n)$ with $x_n=y~\forall~n\in\mathbb{N}$. Then it is $\lim_n x_n=y$. So a sequence in $\mbox{rn}(f)$ is found that converges to $y\in Y$.

For $T$ it's the same proof.

Am I right?

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    If $T$ is bijective then its range is $C[0,1]$. Since $C[0,1]$ is dense in itself, $T$ has dense range. – AGM Mar 30 '14 at 15:46
  • @AGM You are right. I made it too complicated using the sequence-criterion. But is my proof at least okay? –  Mar 30 '14 at 15:47
  • Yes, you proved that every point in the codomain is the limit of a sequence in the range. For metric spaces this implies that the closure of the range is the entire codomain, so the range is dense. – AGM Mar 30 '14 at 15:50

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