My Math Forum does open in M imply open in A?

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 October 31st, 2011, 06:30 PM #1 Senior Member   Joined: Nov 2009 Posts: 169 Thanks: 0 does open in M imply open in A? Suppose that A is open in (M,d) and that G is subset of A. Show that G is open in A if and only if G is open in M. Is this still true if we replaced open everywhere by closed? I have proved the one directon, suppose G is open in A. Then for all x in A, there exists$\epsilon$$>0$such that $B^A_{\epsilon}(x)$$\subseteq$ G. As A is open in (M,d) then for all x in M, there exists $\epsilon$$>0$ such that $B^M_{\epsilon}(x)$$\subseteq$$B^A_{\epsilon}(x)$$\subseteq$ G $\subseteq$ A. Therefore G is open in M. how do i prove the other direction, is this still ture if its closed?

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