I want to prove the equivalence of the following two definitions:

  1. A set $\mathcal B\subseteq\mathcal T$ is called a topology over $\Omega$ if and only if for every $O\in\mathcal T$, there exists a $\mathcal B'\subseteq\mathcal B$ such that $O = \bigcup_{B'\in\mathcal B'}B'$.

  2. A set $\mathcal B\subseteq 2^\Omega$ is called a basis for a topology $\mathcal T$ if and only if for every finite $\mathcal B'\subseteq\mathcal B$, there exists a $\mathcal B''\subseteq\mathcal B$ such that $\bigcap_{B'\in\mathcal B'}B' = \bigcup_{B''\in\mathcal B''}B''$. (using the convention $\bigcup_{i\in\emptyset}i=\emptyset$ and $\bigcap_{i\in\emptyset}i=\Omega$)

The "$\Rightarrow$" direction is pretty easy: For any finite $\mathcal B'\subseteq\mathcal B$ it holds that $\bigcap_{B'\in\mathcal B'}B'\in\mathcal T$ (since $\mathcal B\subseteq\mathcal T$ and $\mathcal T$ is a topology). But since $\mathcal B$ is a basis, it follows 2.

I struggle with the "$\Leftarrow$" direction. I googled and found the following proof idea: Put $\mathcal T := \{\bigcup_{V\in\mathcal V}V:\mathcal V\subseteq\mathcal V\}$ and show that this is indeed a topology. $\emptyset,\Omega\in\mathcal T$ is easy using the conventions. It remains to show that for $\mathcal A\subseteq\mathcal T$ we have that $\bigcup_{A\in\mathcal A}A\in\mathcal T$ and for finite $\mathcal A\in\mathcal T$ we have that $\bigcap_{A\in\mathcal A}A\in\mathcal T$.

For union I had the following idea: Let $\mathcal A\subseteq\mathcal T$. By construction of $\mathcal T$, ther exists a $\mathcal V_A\subseteq\mathcal B$ such that $A = \bigcup_{V_A\in\mathcal V_A}V_A$ for every $A\in\mathcal A$. Taking unions yields $$\begin{align*}\bigcup_{A\in\mathcal A}A &= \bigcup_{A\in\mathcal A}\bigcup_{V_A\in\mathcal V_A}V_A\\ &= \bigcup_{W\in\bigcup_{A\in\mathcal A}\mathcal V_A}W\end{align*}$$ and since $\bigcup_{A\in\mathcal A}\mathcal V_A\subseteq\mathcal B$ it follow that $\bigcup_{A\in\mathcal A}A\in\mathcal T$. However, I am not really sure wether it was okay how I treated the indices.

For the intersection I had no idea. I could show it with induction but that made the proof so lengthy. Does someone have an idea wether there is way to show it in a similar way as the union or at least without induction?

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  • 2
    $\begingroup$ What you did so far looks good to me. You might want to work on the formulation of the two definitions, right now, the first claims to be a definition of a topology even though it is the definition of a basis of a topology. Also, note that the two definitions are different in the sense that the first needs the topology to be given and the second constructs it out of the basis. Thus, you need to make appropriate changes in the formulation so that they actually become equivalent definitions. For the intersection, it might be easier to intersect only two elements of the topology. $\endgroup$ – Matthias Klupsch Feb 28 '18 at 11:24

For intersection note that $$(\bigcup_{V\in\mathcal V_1}V)\cap(\bigcup_{W\in\mathcal V_1}V)=\bigcup_{V\in\mathcal V_1,W\in\mathcal V_2}(V\cap W)\tag1$$

Here $V\cap W=\bigcap_{B'\in\mathcal B'}B' = \bigcup_{B''\in\mathcal B''}B''$ for $\mathcal B'=\{V,W\}\subseteq\mathcal B$.

So every intersection $V\cap W$ in $(1)$ can be replaced by a union of elements of $\mathcal B$.

That implies that the LHS of $(1)$ is an element of $\mathcal T$.

If the intersection concerns more than $2$ sets then you can just repeat.

Side note:

I respected your notation, but can be done much nicer.

Observe that e.g. $\bigcup_{V\in\mathcal V_1}V$ can be written as $\bigcup\mathcal V_1$.

This because $$\bigcup A:=\{x\mid\exists a\in A[x\in a]\}=\bigcup_{a\in A}a$$

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Let $\mathcal{V}_1, \ldots, \mathcal{V}_n$ be subsets of of $\mathcal{B}$.

\begin{align}S =\left(\bigcup_{V_1 \in \mathcal{V}_1} V_1\right) \cap \left(\bigcup_{V_2 \in \mathcal{V}_2} V_2\right) \cap \cdots \cap \left(\bigcup_{V_n \in \mathcal{V}_1} V_n\right) &= \bigcup_{V_1 \in \mathcal{V_1}}\cdots \bigcup_{V_n \in \mathcal{V_n}} V_1 \cap \cdots \cap V_n \end{align}

Now, because of $(2)$, each $V_1 \cap \cdots \cap V_n$ is in fact equal to a union of $\mathcal{B}_{V_1, \ldots, V_n} \subseteq \mathcal{B}$:

$$V_1 \cap \cdots \cap V_n = \bigcup_{B \in \mathcal{B}_{V_1, \ldots, V_n}} B$$

Now we have:

$$S = \bigcup_{V_1 \in \mathcal{V_1}}\cdots \bigcup_{V_n \in \mathcal{V_n}} V_1 \cap \cdots \cap V_n = \bigcup_{V_1 \in \mathcal{V_1}}\cdots \bigcup_{V_n \in \mathcal{V_n}} \bigcup_{B \in \mathcal{B}_{V_1, \ldots, V_n}} B \in \mathcal{T} = \bigcup_{V_1 \in \mathcal{V}_1, \ldots, V_n \in \mathcal{V_n}, B \in \mathcal{B}_{V_1, \ldots, V_n}} B$$

Therefore, $\mathcal{T}$ is a topology and $\mathcal{B}$ is a basis for $\mathcal{T}$.

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