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Let $(E,\tau_1)$ and $(F,\tau_2)$ be Hausdorff locally convex topological vector spaces with topologies $\tau_1$ and $\tau_2$, respectively. Let $T:(E,\tau_1)\longrightarrow (F,\tau_2)$ be a topological isomorphism, that is, $T$ is linear, bijective and continuous with continous inverse $T^{-1}$.

Let $M\subset E$ and $N\subset F$ be subspaces such that the restriction $T|_M:M\longrightarrow N$ is an algebraical isomorphism, that is, $T|_M$ is linear and bijective.

Is it true that $(\overline{M}^{\tau_1},\tau_1)$ is topologically isomorphic to $(\overline{N}^{\tau_2},\tau_2)$ via the mapping $T$ ? (where $\overline{M}^{\tau_1}$ is $\tau_1$-closure of $M$ in $E$ and $\overline{N}^{\tau_2}$ is $\tau_2$-closure of $N$ in $F$.)

I guess (intuitively) the answer is "yes", but how to see this fact ? Any hint/comment/answer will be appreciated.

Paul Frost
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serenus
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    It's enough (why?) to show the image of $\overline{M}$ is $\overline{N}$, which follows from $T$ being a homeomorphism. – Wojowu Feb 09 '19 at 22:16

1 Answers1

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It is true. The vector space structure of $E, F$ is completely irrelevant. If you have a homeomorphism $f : X \to Y$ between topological spaces $X,Y$ (in your question: $T : E \to F$), then for any subset $M \subset X$ restriction produces a homeomorphism $f_M : M \to f(M)$. Since $f$ is a homeomorphism, you have $f(\overline{M}) = \overline{f(M)}$ so that $f_{\overline{M}} : \overline{M} \to \overline{f(M)}$ is a homeomorphism.

Paul Frost
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