# Combining two topological spaces so that one is dense.

Let $X$ and $Y$ be topological spaces. For fun let's assume they are metric. Is there a topology on $X\cup Y$ (disjoint union) so that $X$ is dense, and the subspace topologies on $X$ and $Y$ are the same as their original topologies?

My attempt: Let $U$ be open in $X\cup Y$ if: (i) $U$ is open in $X$, or (ii) $U=V\cup W$ for some nonempty open $V\subseteq X$ and $W\subseteq Y$.

Everything I wanted is satisfied, except that it is not a topology because the intersection of two of these new open sets is not necessarily a new open set.

EDIT: Someone gave an example below, which is $T_0$. Now I want a $T_1$ topology. Is that possible?

Sure - $U$ is open iff

• $U$ is an open subset of $X$ - in particular, $U$ is disjoint from $Y$; or

• $U$ is of the form $X\cup V$ for some open subset of $Y$.

That is, as soon as an open set has anything from $Y$, it must contain all of $X$.

(I'm of course assuming that $X$ and $Y$ are disjoint.)

EDIT: We can get a $T_1$ example, assuming at least one of $X$ or $Y$ is sufficiently "big":

Suppose $X$ and $Y$ are topological spaces, and there is a family of closed sets $\mathcal{I}$ on $X$ such that

• $\mathcal{I}$ is closed under finite unions and arbitrary intersections,

• for each $x\in X$ there is $C\in \mathcal{I}$ such that $x\not\in C$, and

• $X\not\in\mathcal{I}$.

(Think of $\mathcal{I}$ as the class of compact sets, although of course it might not be exactly that.) Then we can form a $T_1$ topology on the disjoint union of $X$ and $Y$, of which $X$ is a dense subset, as follows: $U\subseteq X\cup Y$ is open iff

• $U\subseteq X$ and $U$ is open in the sense of $X$, or

• $U$ is of the form $B\cup V$ for some open $V\subseteq Y$ and some $B\subseteq X$ with $X\setminus B\in\mathcal{I}$.

In particular, this works if every point in $X$ is closed and $\vert X\vert\ge \aleph_0$ (set $\mathcal{I}$ to be the set of finite subsets of $X$ - in notation, $\mathcal{I}=[X]^{<\omega}$).

• Note that this is asymmetric, in that $X$ and $Y$ are treated differently. – Noah Schweber Dec 15 '15 at 0:23
• Ok, now how about one that is $T_1$? – Forever Mozart Dec 15 '15 at 0:26
• @ForeverMozart Getting a $T_1$ space is already impossible if $X$ and $Y$ are one-point spaces. – Slade Dec 15 '15 at 0:34
• @Slade Oh, duh! – Forever Mozart Dec 15 '15 at 0:35
• @ForeverMozart See my edit. – Noah Schweber Dec 15 '15 at 2:07