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Let $M$ be a metric space and let $X$ be a subset of $M$ with the relative metric. If $Y$ is a subset of $X$, let $\overline{Y}^{(X)}$ denote the closure of $Y$ in the metric space $X$. Prove that $\overline{Y}^{(X)}=\overline{Y}\cap X$. State and prove a corresponding result for $Y^o$.

Suppose $y\in \overline{Y}^{(X)}$. Since $\overline{Y}^{(X)}$ is defined in the metric space $X$, we must have $y\in X$. Also, there exists a sequence $y_1,y_2,\ldots\in Y$ converging to $y$. So $y\in \overline{Y}$, meaning that $Y\in \overline{Y}\cap X$. On the other hand, suppose $Y\in \overline{Y}\cap X$. Then there exists a sequence $y_1,y_2,\ldots\in Y$ converging to $y$ in the metric space $M$. Since $Y\subset X$, that same sequence also shows that $y$ is a limit point in the metric space $X$.

Also, $Y^{o(X)} = Y^o\cap X$. Since $Y\subset X$, any interior point of $Y$ in the metric space $X$ must be an interior point in the metric space $M$, and vice versa.

I feel like this is too straightforward... am I misunderstanding something?

Edit: Okay I see where I went wrong. Let me try again on the interior points.

\begin{align*}Y^{o(X)} &= \bigcup\{Z\mid Z\subset Y, Z \text{ open in } X\} \\ &=\bigcup\{Z\cap X\mid (Z\cap X)\subset Y, Z \text{ open in } M\} \\ &=\bigcup\{Z\cap X\mid Z\subset (Y\cup X'), Z \text{ open in } M\} \\ &=\bigcup\{Z\mid Z\subset (Y\cup X'), Z \text{ open in } M\} \cap X \\ &=(Y\cup X')^o\cap X \end{align*}

Can we simplify $(Y\cup X')^o$ further? Does the $^o$ distribute inside?

Edit 2 Seems like the answer to the line above is no :)

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  • $\begingroup$ I think you meant $\ ^\circ$ on the last left hand side above? What is $X'$? $\endgroup$ – copper.hat Jun 17 '13 at 15:52
  • $\begingroup$ Edited the $^o$. And $X'$ is my notation for the complement of $X$, i.e. $M- X$ (what notation do you usually use?) $\endgroup$ – PJ Miller Jun 17 '13 at 15:55
  • $\begingroup$ Okay $(A\cup B)^o=A^o\cup B^o$ is not true, for example $A=(0,1],B=(1,2)$. So this is perhaps the simplest form. $\endgroup$ – PJ Miller Jun 17 '13 at 15:58
  • $\begingroup$ $X'$ is reasonable for complement (I use $X^c$). $\endgroup$ – copper.hat Jun 17 '13 at 16:05
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The closure argument looks good, maybe one thing to be a little more careful about is that when you say a sequence converges, say it converges as a sequence in the desired space (differentiating between convergence in $X$ and in $M$. You started doing this towards the end, but the beginning could use it. But this is really just me searching really hard for ways to improve).

Here is another way of doing it that doesn't rely on the use of sequences (this is useful when working with an arbitrary topological space). One characterization of closure is the following:

Let $A$ be a topological space (e.g. a metric space), and $B\subset A$. Then $\overline{B} = \bigcap\{Z|Z\supset B, Z \text{ closed in } A\}$. That is to say, $\overline{B}$ is the smallest closed set containing $B$.

So \begin{align*}\overline{Y}^{(X)} &= \bigcap\{Z|Z\supset Y, Z \text{ closed in } X\} \\ &=\bigcap\{Z\cap X|Z\supset Y, Z \text{ closed in } M\} \\ &=\bigcap\{Z|Z\supset Y, Z \text{ closed in } M\} \cap X \\ &=\overline{Y} \cap X. \end{align*}

Edit: Wow, I was way off on the relative interior bit. Thank you @copper.hat for your example.

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  • $\begingroup$ Thanks, please see my edit! $\endgroup$ – PJ Miller Jun 17 '13 at 15:50
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The formula does not hold for the interior.

Let $M=[0,2]$ with the usual distance, $X=[0,1]$, $Y=(\frac{1}{2},1]$.

Then $Y$ is open in $X$ (since $Y=(\frac{1}{2},\frac{3}{2}) \cap X$), hence $Y^{\circ^{(X)}} = Y$, but $Y^\circ = (\frac{1}{2},1)$.

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  • $\begingroup$ Thanks for your example. Please see my edit! $\endgroup$ – PJ Miller Jun 17 '13 at 15:50

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