This was a question that I had while revising metric spaces.
For a metric space $X$, let $A\subset X$. We know that if $G, H$ are disjoint open sets in $A$, $\exists$ disjoint open sets $U, V \subset X$ such that $G = U \cap A$, $H = V \cap A$.
What I was wondering is that if we can generalize it to general topological spaces, that is, for a topological space $X$ and $A$ having subspace topology.
Also, if this does not hold for a general topological space, what restrictions (like Hausdorff-ness) do we need to place for this to hold?
Best Answer
You don't have it in general, a simple example where things go wrong: $X=\Bbb R$ in the included point topology wrt $0$: $$\mathcal{T}=\{O \subseteq \Bbb R: 0 \in O \lor O = \emptyset\}$$
Then $A = \Bbb R \setminus \{0\}$ is discrete as a subspace (so $U=\{1\}$ and $V=\{2\}$ are disjoint and open in $A$), but no non-empty subsets in $X$ are disjoint. The unique subsets that give $U$ and $V$ as intersection are $U \cup \{0\}, V \cup \{0\}$ which intersect in $0$..
This example is $T_0$ but not $T_1$. A finite $A$ in an infinite cofinite space is another example: it's discrete but the whole space is hyperconnected, all non-empty open sets intersect.
The Sorgenfrey plane is a Tychonoff counterexample: $U=\{(x,-x): x \in \Bbb Q\}$ and $V=\{(x,-x): x \in \Bbb R\setminus \Bbb Q\}$ are disjoint open in $A=\{(x,-x): x \in \Bbb R\}$ (which is closed and discrete) but do not have disjoint open subsets in the whole "plane" (see Counterexamples in Topology for an argument), or (more abstractly) note that there are disjoint sets in $A$ without open disjoint neighbourhoods (by Jones' lemma argument) and that these will also suffice. More Tychonoff examples like this exist (Mrówka $\Psi$-space, the rational sequence topology etc. involving similar arguments. Maybe examples with better separation axioms can be found too.