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If $Y$ is a subset of topological vector space $X$ and is compact and convex show that $\overline{Y^\circ} = \overline{Y}$ and $\overline{Y}^\circ = Y^\circ$.

I tried this way but I am not sure:

$Y$ is compact so $Y = \overline{Y}$. Then it follows that $Y^\circ = (Y)^\circ = (\overline{Y})^\circ$. And for another I stuck so could some help.

One more thing: What happens if $Y$ is not compact and convex the proof is true for this case too.

This is not a homework question.

  • 0
    Y is compact. This does NOT imply that Y is closed, so your atempt to proof this theorem is not ok from the starting point on.2012-11-30
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    @Fant could you help me to solve this2012-11-30
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    @motu: Is a topological vector space assumed to be Hausdorff?2012-11-30
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    @NilsMatthes what u mean2012-11-30
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    A topological vector space is, in almost all applications, assumed to be Hausdorff, so in that case $Y$ compact does imply $Y$ closed, if I recall my general topology right.2012-11-30
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    @ChristopherA.Wong so could you show me a way2012-11-30
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    @motu: Which part of my question is not clear? Do you know what "Hausdorff" means in the context of topological spaces? The reason I ask this is that in a Hausdorff space every compact set is closed, as was already remarked by Christopher.2012-11-30
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    $[0,1]\times\{0\}$ is a compact and convex subset of $\mathbb R^2$, but its interior is empty.2012-12-01
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    @JonasMeyer Game over. Or $Y=\{0\}$ in $\mathbb R$.2012-12-01
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    @NilsMatthes of course but still i am not clear but i found this is true from prop 1.2.6 from pedersen's analysis now. any way can some one give me the solution2012-12-01

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