I cannot understand that $\mathfrak{O} := \{\{\}, \{1\}, \{1, 2\}, \{3\}, \{1, 3\}, \{1, 2, 3\}\}$ is a topology on the set $\{1, 2, 3\}$.

Question

I cannot understand that $\mathfrak{O} := \{\{\}, \{1\}, \{1, 2\}, \{3\}, \{1, 3\}, \{1, 2, 3\}\}$ is a topology on the set $\{1, 2, 3\}$.

Let $\Lambda$ be any set, and $O_{\lambda} \in \mathfrak{O}$ for all $\lambda \in \Lambda$.

Then, we must verify the following: $$\bigcup_{\lambda \in \Lambda} O_{\lambda} \in \mathfrak{O}.$$

If $\Lambda$ is a finite or countable set, then $$\bigcup_{\lambda \in \Lambda} O_{\lambda} \in \mathfrak{O}$$ is obvious because $O_1 \cup O_2 \in \mathfrak{O}$ for any $O_1, O_2 \in \mathfrak{O}$.

But if $\Lambda$ is not countable, I think $$\bigcup_{\lambda \in \Lambda} O_{\lambda} \in \mathfrak{O}$$ is not obvious. Of course, $$\bigcup_{\lambda \in \Lambda} O_{\lambda} \in 2^{\{1, 2, 3\}}$$ is obvious.

Please tell me a formal proof.

Answer 1

The definition of a topology says that “the union of any collection $\Lambda$ of open sets is an open set,” where “$X$ is an open set” means $X\in\mathfrak O.$ When one is learning about topology this is usually clarified by noting that this may be a union of countably many sets or indeed of uncountably many sets. This observation is not part of the definition.

Nevertheless, let us prove that any union of sets in $\mathfrak O$ as given in your question is also in $\mathfrak O$. First observe that for any set $X$ we have $X\cup X=X$ and so $\bigcup_{\lambda\in\Lambda} X_\lambda = \bigcup_{\lambda\in\Lambda’}X_\lambda$ where $\Lambda’$ is a set of unique representatives of the equivalence classes of $\Lambda$ under the relation $\lambda\sim\mu$ iff $X_\lambda = X_\mu$. To prove this, suppose $x\in\bigcup_\Lambda X_\lambda$ Then $x\in X_\lambda$ for some $\lambda$ and the equivalence class $[\lambda]$ has some unique representative $\mu\in\Lambda’$ and $\lambda\sim\mu$ so $X_\lambda=X_\mu$ so $x\in X_\mu$ so $x\in\bigcup_{\lambda\in\Lambda’}X_\lambda.$ Thus we get LHS $\subseteq$ RHS and because $\Lambda’\subseteq\Lambda$, we get RHS $\subseteq$ LHS.

Now $\Lambda’$ May only contain as many different values as there are different values for $X_\lambda$ and as in your case $X_\lambda\in\mathfrak O,$ you must have $\Lambda’$ finite. Therefore the “uncountable” case (which isn’t really uncountable at all) may be reduced to the finite case which presumably you can solve yourself.

Answer 2

If $\Lambda$ is not countable, there are many duplicates among the $O_\lambda$, since $\mathfrak{O}$ itself is finite.

So $\{ O_\lambda \mid \lambda \in \Lambda\}$ will always be finite, since it is a subset of $\mathfrak{O}$. Note that the only thing that you really have to verify is that $O \cup O' \in {\mathfrak O}$ for every $O, O' \in {\mathfrak O}$.