2

Assume we have two topological spaces $(X, \mathcal{O}_1)$ and $(X, \mathcal{O}_2)$ with two subbases $\mathcal{S}_1$ and $\mathcal{S}_2$, resp.

Now I asked myself whether the following statement holds:

If $\mathcal{S}_1 \subseteq \mathcal{O}_2$ and $\mathcal{S}_2 \subseteq \mathcal{O}_1$, then we have $\mathcal{O}_1 = \mathcal{O}_2$

I think I proved it, but since I'm new to topology I'm not quite sure whether my argument is correct. My idea:

Since $\mathcal{S}_1$ is a subbase of $\mathcal{O}_1$, $\mathcal{S}_1 \cup \mathcal{S}_2 \supseteq \mathcal{S}_1$ is also a subbase of $\mathcal{O}_1$.

By the same argument $\mathcal{S}_1 \cup \mathcal{S}_2 \supseteq \mathcal{S}_2$ is a subbase of $\mathcal{O}_2$.

So $\mathcal{O}_1$ and $\mathcal{O}_2$ have a common subbase, therefore they must be equal.

Edit: also, can we show the same thing if we have two spaces $X$ and $Y$ and a priori don't know if $X=Y$? Because I think this follows from $X\in \mathcal{O}_1\cap \mathcal{O}_2$ and $Y \in \mathcal{O}_1\cap \mathcal{O}_2$.

Community
  • 1
Jakob W.
  • 860
  • 1
    The answer is yes: Finite intersection of elements in a subbase gives you basis elements and since finite intersection are again in the topology, you actually get inclusions of bases. – noctusraid Mar 12 '19 at 12:28
  • The question in your new edit makes little sense, I'm not even sure what you intended to ask. If your intention is that $\mathcal O_1$ is a topology on $X$ and $\mathcal O_2$ is a topology on $Y$, then the answer varies wildly depending how much $X$ and $Y$ overlap. One thing I can say for sure: if $X$ and $Y$ are disjoint sets, then $\mathcal O_1 \cap \mathcal O_2 = {\emptyset}$. – Lee Mosher Mar 12 '19 at 14:07

1 Answers1

1

$\mathcal{S}_1$ is a subbase for $\mathcal{O}_1$, which means that $\mathcal{O}_1$ is the smallest topology on $X$ that contains $\mathcal{S}_1$ as a subset.

If now $\mathcal{S}_1 \subseteq \mathcal{O}_2$, that minimality of $\mathcal{O}_1$ (as $\mathcal{O}_2$ is a topology that contains $\mathcal{S}_1$) gives us that $$\mathcal{O}_1 \subseteq \mathcal{O}_2$$ and the reverse inclusion follows in the same way, mutatis mutandis.

So we indeed have equality of topologies.

Henno Brandsma
  • 242,131