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Nov 6, 2011 at 20:56 comment added Peter Where can I find a reference about it? Or better, how do you prove that fact? (I mean, $Y\le X$ implies $dc(Y)\le dc(X)$)?
Nov 5, 2011 at 14:01 comment added Ramiro de la Vega Peter $dc(\mathbb{I})=\aleph_0$ since it is in fact second countable. The claim says "any metric space" and "any subspace".
Nov 5, 2011 at 13:38 vote accept Peter
Nov 5, 2011 at 13:38 vote accept Peter
Nov 5, 2011 at 13:38
Nov 4, 2011 at 20:00 comment added Peter Take $X:=\mathbb{R}$ and $Y:=\mathbb{I}$ with the usual metric. Notice that $dc(X)=\aleph_0$ and $dc(Y)=2^{\aleph_0}$. Does your claim hold if both $X$ and $Y$ are complete in the metric sense?
Nov 3, 2011 at 21:28 history answered Ramiro de la Vega CC BY-SA 3.0