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Let $O$ be a nonempty open subset of a separable (Lindelöf) topological space $( X,\tau)$. Prove that $O$ as a subspace of $X$, is also separable (Lindelöf).

(1-separable) Given a countable dense subset of $X$, if I intersect with the open subset of $X$, we get a countable dense subset of the open subset.

(2-Lindelöf) Let $U$ be an open covering of the subspace $O$. Since all the elements of $U$ are open are open $O$, they equal the intersection of some family of open sets with $X$, call it $U'$.

I am kind of in the clouds about this proof. Please help me. Thank you! Klara

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  1. Let $D$ a countable dense subset of $X$. You have to show that $D\cap O$ is dense in $O$. Let $x\in O$ and $V$ a neighborhood of $x$ in $O$. We can assume that $V=O\cap O'$, where $O'$ is an open subset of $X$. Since $O\cap O'$ is a neighborhood of $x$ in $X$ and $D$ is dense in $X$, there is $y\in D\cap O\cap O'$, so $x$ is in the closure of $D\cap O$ for the induced topology.

  2. You will be interested by this.

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    Im lost. $T$hank you though.2012-11-27
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There exists a Lindelöf space $X$ with an open subspace $O$ which is not Lindelöf.

Let $X=[0,\omega_1]$. It is compact, and hence it is Lindelöf. Let $O=\{x\in X: x \text{ is isolated in } X \}$. It is not difficult to see $O$ is an uncountable open subspace of $X$. The cover $\mathscr U=\{\{x\}: x\in O\}$ of $O$ has no countable subcover, which witnesses that $O$ is not Lindelöf.