Let $(X, \tau_1)$ and $(X,\tau_2)$ be two topological spaces having the same underlying set X.
Let $\tau$ be the smallest topology on $X$ such that identity maps $I_1 : (X,\tau) \rightarrow (X,\tau_1)$ and $I_2 : (X,\tau) \rightarrow (X,\tau_2)$ are continuous.

Then if both $(X,\tau_1)$ and $(X,\tau_2)$ are separable topological space. Does it imply that $(X,\tau)$ will be separable?


It does not; here’s a counterexample.

Let $Y=\{y_k:k\in\Bbb N\}$ and $Z=\{z_k:k\in\Bbb N\}$ be disjoint countably infinite sets. Two subsets of $\Bbb N$ are said to be almost disjoint if their intersection is finite; let $\mathscr{D}$ be an uncountable family of almost disjoint subsets of $\Bbb N$. (Two constructions of such a family can be found in this answer.) Let $X=Y\cup Z\cup\mathscr{D}$. Let

$$\mathscr{I}=\big\{\{y_k\}:k\in\Bbb N\big\}\cup\big\{\{z_k:k\}\in\Bbb N\big\}\;.$$

For each $D\in\mathscr{D}$ let $D_Y=\{y_k:k\in D\}$ and $D_Z=\{z_k:k\in D\}$, and let

$$\mathscr{B}_1(D)=\{D\}\cup\{D_Y\setminus F:F\subseteq Y\text{ and }F\text{ is finite}\}$$


$$\mathscr{B}_2(D)=\{D\}\cup\{D_Z\setminus F:F\subseteq \text{ and }F\text{ is finite}\}\;.$$



$\mathscr{B}_1$ and $\mathscr{B}_2$ are bases for topologies $\tau_1$ and $\tau_2$, respectively, on $X$. Both topologies are separable: $Y\cup Z$ is dense in both.

The coarsest topology $\tau$ on $X$ making the identity maps from $\langle X,\tau\rangle$ to $\langle X,\tau_1\rangle$ and $\langle X,\tau_2\rangle$ continuous is the joint of $\tau_1$ and $\tau_2$, i.e., the topology generated by the subbase $\tau_1\cup\tau_2$, which has as a base the sets of the form $U\cap V$ with $U\in\tau_1$ and $V\in\tau_2$. Clearly $\mathscr{I}\subseteq\tau$. Let $D\in\mathscr{D}$ be arbitrary, and let $U=\{D\}\cup Y$ and $V=\{D\}\cup Z$; then $U\in\tau_1$ and $V\in\tau_2$, so $\{D\}=U\cap V\in\tau$. Thus, $\tau$ is the discrete topology on the uncountable space $X$, and $\langle X,\tau\rangle$ is not separable.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.