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I was wondering if you could help me with this proof I have for metric spaces. I understand the theorem and the idea I just don't understand how to proceed. I am attaching the question.

Here is how I think I should proceed: So, Let U = Q(i.e. rational numbers), Y = [0,1], and X = R(i.e. real numbers).

Am I doing it right?

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  • $\begingroup$ Set of rationals is neither open nor closed set in R $\endgroup$ – Curious Oct 30 '17 at 8:21
  • $\begingroup$ For any Y not open in X let U=Y. $\endgroup$ – DanielWainfleet Oct 30 '17 at 12:08
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Assuming you meant $U = \mathbb{Q} \cap Y$, your example does not work, since $U$ is both not open in $Y$ and not open in $X$.

A counterexample can be made with $U=Y = \{0\}$ in $X=\mathbb{R}$.

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Consider the metric space $X=\mathbb{R}$ and the closed subset $Y=[0,1]$. What are the open sets in $Y$? Are they all open in $X$?

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