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If I have open sets $U$ and $V$ in a space $X$ $$U \subset V \subset X,$$ does it imply that the closure of $U$ is still included in $V$?

If yes, I would like to know why and if not, what would be sufficient conditions for this to hold.

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    What do you know about the closure of a set? – 2017-01-29
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    that it's the smallest closed set in which the original set is included – 2017-01-29
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    What if $U=V$? ${}{}$ – 2017-01-29
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    with the subset sign i meant that it's strictly included, so they are not equal – 2017-01-29
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    Sorry, i have misread your question, i thought you asked if closure of U is contained in the closure of V, my bad. – 2017-01-29

2 Answers 2

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$(0,1)\subset (0,2)$ but $[0,1]$ the closure of $(0,1)$ is not contained in $(0,2)$.

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    ok, do you maybe know something about the sufficient conditions? – 2017-01-29
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Simplest way to see why it is false is to take $U=V$