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This is the claim to prove:

If the open sets $G_Y'$ and $G_Y''$ in $Y$ do not intersect, then the corresponding open sets $G_X'$ and $G_X''$ in $X$ can be choosen so that they also have no points in common.

The existence of two open sets $ G_X '$ and $ G_X' '$ containing $ G_Y' $ and $ G_Y '' $ respectively is trivial to prove: $ G_Y' $ open in $ (Y, d ') \Rightarrow $ for every point $ y' \in G_Y' $ there is a ball $ B_Y (y', r') \subset G_Y' \Rightarrow $ the set of the corresponding balls $ B_X (y', r') $ in $ (X, d) $ is a set that contains $ G_Y' $ and which is open since union of open.

Then, as regards the actual statement, I did so: given $ G_Y '$ and $ G_Y''$, we build the set $ G_X' $ as described above, with the foresight to choose each ball $ B_Y (y', r') $ so that it never contains points of $ G_Y '' $ (which is always possible since $ G_Y '$ open and $ G_Y' \cap G_Y '' = \emptyset $). Having done this, we continue with the construction of the set $ G_X '' $: for each point $ y '' \in G_Y '' $ we consider a ball $ B_Y (y '', r '') $ of sufficiently small radius so such that the corresponding ball in $ X $ does not contain points of $ G_X '$ (which is always possible, because otherwise it would be that every neighborhood of $ y' '$ in $ X $ would contain points of $ G_X' \Rightarrow $ there will be a ball $ B_X (y '', \widetilde{r} '') $ which will only contain points of $ G_X '\setminus G_Y' $ because otherwise every neighborhood of $ y '' $ in $ X $ would always contain points of $ G_Y '$ and therefore $ y' '$ would be a limit point for $ G_Y' $, which is absurd by hypothesis; removing the points $ \{x' \in G_X '| d (x', y' ') \leq \widetilde{r} '' /2 \} = \widetilde{B}_X (y '', \widetilde{r} '' / 2) \cap G_X '$ from $ G_X' $ you get a new $ \widetilde{G}_X '$ open which still contains $ G_Y' $). Therefore, by construction $ G_X '\cap G_X' '= \emptyset $.

I'm pretty sure it could have been done more elegantly. Is it true?

Nameless
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2 Answers2

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A classic lemma on metric subspaces (useful in dimension theory, e.g. see van Mill, infinite-dimensional topology, prerequisites and an introduction p. 127):

let $Y$ be a subspace of the metric space $X$ and denote by $\rho Y$ ($\rho X$) the set of closed subsets of $Y$ (resp. $X$) and define $d(x,\emptyset)=+\infty$ for convenience. Then for a closed $A \subseteq Y$ the set $\kappa(A)=\{x \in X: d(x,A) \le d(x,Y \setminus A) \}$ defines a map from $\rho Y \to \rho X$ with the following properties

  1. $\kappa(\emptyset)=\emptyset, \kappa(Y)=X$.
  2. $\kappa(A) \cap Y = A$ for any $A \in \rho Y$.
  3. If $A \subseteq B, A,B \in \rho Y$ then $\kappa A \subseteq \kappa(B)$.
  4. If $A,B \in \rho Y$, $\kappa(A \cup B)=\kappa(A) \cup \kappa(B)$.

The proof isn't hard.

Now, if you have disjoint open $O, O'$ in $Y$. Then $A := Y\setminus O$ is in $\rho Y$, and so is $B:= Y \setminus O'$ and disjointness tells us that $A \cup B = Y$ so $\kappa(A) \cup \kappa(B)=\kappa(Y)=X$ and so it's clear that $U = X\setminus \kappa(A)$ and $U' = X\setminus \kappa(B)$ are the required disjoint "$X$-extensions" for $O$ resp. $O'$.

Henno Brandsma
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  • Kuratowski's idea (1948) is the basis for this, and is an exercise in Engelking: extend an open set $O$ in $Y$ by $\hat{O}:={x \in X: d(x,O) < d(Y\setminus O}$ to avoid the complements. If finitely many open sets have empty intersection in $Y$, their corresponding extensions also have. Same trick as the above lemma, really. – Henno Brandsma May 31 '20 at 08:40
  • The Kuratowski paper gives full proofs (in French, but it's easy enough to follow). – Henno Brandsma May 31 '20 at 12:59
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Suppose $Y$ is a subspace of a metric space $X$, and $U_1,U_2$ are subsets of $Y$ which are open in $Y$. Assuming $U_1$ and $U_2$ are nonempty, the disjoint sets $$V_1=\{x\in X:d(x,U_1)\lt d(x,U_2\}$$ and $$V_2=\{x\in X:d(x,U_2)\lt d(x,U_1\}$$ are open because the functions $x\mapsto d(x,U_1)$ and $x\mapsto d(x,U_2)$ are continuous. Finally, $U_1\subseteq V_1$ because $$x\in U_1\implies d(x,U_1)=0\lt d(x,Y\setminus U_1)\le d(x,U_2)\implies x\in V_1,$$

and similarly $U_2\subseteq V_2$.

That's assuming $U_1$ and $U_2$ are nonempty. Of course, if $U_i=\emptyset$, we can simply take $V_i=\emptyset$ and $V_{3-i}=X$.

P.S. We may as well let $Y=U_1\cup U_2$. That is, all we need to assume is that $U_1$ and $U_2$ are disjoint sets which are open in $U_1\cup U_2$, which is another way of saying that $U_1$ and $U_2$ are separated sets, disjoint sets such that neither contains a limit point of the other. We have shown that in a metric space, any two separated sets (for example, disjoint closed sets) are contained in disjoint open sets.

bof
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