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We assume that $X$ is a topological vector space with a topology $\tau$. I want to show that if $F$ is a closed subset of $X$, then its translate $a+F$ is also closed in $X$. Am I right to say that $$ (a+F)^c=a+F^c?$$ If this is so, then how can we prove it? If that is true, then the result follows because $\tau$ is translation invariant. Alternately, the result can be proved from the fact that if $a\in X$ then the translation map $$ T_a: X \to X$$ defined by $T_a(x)=a+x$ for every $x\in X$ is a homeomorphism and hence a closed map. This means that if $F$ is closed, then $T_a(F)=a+F$ is closed.

juniven

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Yes. $y\in a +F$ if and only if there exists $x\in F$ such that $y = a + x$, while $y\in a+F^c$ if and only if there exists $z\in F^c$ such that $y = a + z$. These conditions simultaneously holding would imply $x=z\in F\cap F^c$. You can use this to see that $y\in (a+F)^c$ if and only if $y\in a+F^c$.

Yes, $T_a$ and $T_{-a}$ are continuous, so $T_a$ is a homeomorphism as you said.

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