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I am currently working through some topology problems, and I would like to confirm my results just for some peace of mind!

Statement: Let $(X, d_x)$ be a metric space, and $Y\subset X$ a non-empty subset. Define a metric $d_y$ on $Y$ by restriction. Then, each subset $A\subset Y $ can be viewed as a subset of $X$.


Q 1: If $A$ is open in $X$, $A$ open in $Y$.

A: I obtained TRUE. Roughly speaking, my proof relied on showing that if we can always draw a neighbourhood in the set $A$ (w.r.t $X$) if we can definitely do so w.r.t $Y$.


Q 2: If $A$ is open in $Y$, $A$ open in $X$.

A: This I found FALSE, example: Let $X =\mathbb{R}, Y = [0,1]$ and $A = [0, 1/2)$. $A$ is open in $Y$, howevever, $A$ is not open in $X$.


Q 3: If $A$ is closed in $X$, then $A$ is closed in $Y$.

A: This I found TRUE. Roughly speaking, given the preliminary conditions, $A$ cannot "lose" any of its limit points when we assess it closure in $Y$.


Q 4: If $A$ is closed in $Y$, then $A$ is closed in $X$.

A: This I found FALSE , example: let $X = \mathbb{R}$, $Y = (0,1]$ and $A = (0,1/2].$ A is closed in $Y$, but not in $X$.

Are any of these deductions wrong? If so, how would you rectify them?

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Apart from any possible errors in the details of carrying out the arguments that you merely sketched, this is fine. –  Brian M. Scott Jan 4 '13 at 17:48

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