$\def\emptyset{\varnothing}$Let $(X, \tau)$ be a topological space. Suppose that $\tau'$ is the collection of open subsets $U \subset X$ with respect to $\tau$ such that either $U=\emptyset$ or $U$ is dense in $X$ with respect to $\tau$.

I want to show that $\tau'$ defines a topology on $X$. To do this, I know that I must show four properties:

1) $\emptyset \in \tau'$

2) $X \in \tau'$

3) If $U_1,U_2, . . ., U_n \in \tau'$, then $\bigcap\limits^n_{i=1} U_i \in \tau'$

4) If $S$ is any nonempty set and for each $s \in S$, $U_s \in \tau'$, then $\bigcup\limits_{s\in S} U_s \in \tau'$.

I was able to complete the first three on my own, but I am struggling with the fourth one. I know that if $U_s \in \tau'$, then either $U_s = \emptyset$ or $U_s$ is dense in $X$ with respect to $\tau$. If $U_s = \emptyset$, it essentially contributes nothing to the union, so you really only need to consider the case where $U_s$ is dense in X with respect to $\tau$. By definition, for every nonempty open set $V \subset X$, $U_s \cap V \neq \emptyset$.

I want to eventually be able to show that $\bigcup\limits_{s\in S} U_s$ is dense in $X$ with respect to $\tau$. Any help on where to go from here would be greatly appreciated.

  • 1
    $\begingroup$ If there is no non-empty open set contained in the complement of $U_s$ then even less in the complement of $\bigcup_{s\in S}U_s$. I would have thought that (3) was the problematic because the intersection gets smaller than the factors. $\endgroup$ – user530511 Feb 11 '18 at 23:58
  • $\begingroup$ Prove 3. For the discrete space, the constructed topology gives the indiscete space. $\endgroup$ – William Elliot Feb 12 '18 at 2:37

Let $V:=\bigcup_{s\in S} U_s$ for ease of notation. We may assume that there is a $t\in S$ such that $U_t\ne\emptyset$ otherwise, $V=\emptyset\in\tau'$. Since $U_t\ne\emptyset$ and $U_t\in\tau'$, we know that $U_t$ must be dense in $X$ by definition of $\tau'$, therefore the closure of $U_t$ is $X$.

Note now that $U_t \subseteq V\subseteq \overline{V}$, so $\overline{V}$ is a closed set containing $U_t$. Therefore, $\overline{V}$ must contain the closure of $U_t$, which is $X$. In summary, we have $X\subseteq\overline{V}\subseteq X$, hence $\overline{V}=X$, which means that $V$ is dense in $X$. Consequently, we have $V\in\tau'$ by definition.

  • $\begingroup$ How do we know that the closure of $U_t$ is $X$? Further, why does this imply that $V$ is dense in $X$? $\endgroup$ – mathqueen459 Feb 12 '18 at 3:15
  • $\begingroup$ @britgirl5 I added some more detail. $\endgroup$ – Jesko Hüttenhain Feb 12 '18 at 12:22

Since $U_{t}$ is dense in $X$, we must have that $\operatorname{Cl}_{X}(U_t) = X$. This is the definition for a subset to be dense in the space $X$. For the latter part of your question, @Jesko Huttenhain has provided a more detailed explanation.


If $A_\alpha$ denote subsets of a space $X,$ we can show in general that $$\overline{\bigcup A_\alpha} \supset \bigcup \overline{A_\alpha}.$$

Here, since $\overline{A_\alpha} = X$ for each $\alpha,$ by the above identity $\overline{\bigcup A_\alpha}=X.$


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.