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I understand that an open set is just a subset of a space that obeys the definition of a topology.

I understand that a basis is a collection of open subsets of a space such that every open set of the space is a union of members of the basis.

But I'm confused for product spaces:

Product spaces $(\prod X_i, T')$ are defined with a basis for $T'$ being generated by $\prod O_i$. So by the definition of a basis, each $\prod O_i$ should be an open set in $(\prod X_i, T')$. But to use a specific case: If $(0,1)\times(0,1)$ and $(2,3)\times(2,3)$ are open sets then their union should be an open set in $R^2$. But this defies the definition of a topology since no $O_1\times O_2$ equals this for $O_1, O_2$ being open subsets in $R$, respectively.

Can someone explain what I'm misunderstanding here?

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    the union of sets from a base is not necessarily in the base. So yes, $(0,1)\times (0,1)\cup(2,3)\times (2,3)$ is a open set but not belongs to the base $O_1\times O_2$ of the product topology – 2017-01-31

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The product topology on $X \times Y$ is specified by the base $$\mathcal{B} = \{O_1 \times O_2: O_1 \subset X \text{ open } ,O_2 \subset Y \text{ open }\}$$

and indeed all open sets of $X \times Y$ are by definition the unions of those basic open sets.

So $(0,1) \times (0,1) \cup (2,3) \times (2,3)$ is indeed open as a union of basic open sets. It's not in $\mathcal{B}$, but a base is not closed under unions, why would it be? Most bases in practice are not closed under unions. You seem to think that $\mathcal{B}$ are all open sets, but there are way many more (e.g. the open circle in the plane is also a union of open squares).

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    I guess I'm not understanding why the set of all sets of the form $(O_1 \times \dots O_n)$ can't be the topology. Because if $(0,1) \times (0,1)$ and $(2,3)\times(2,3)$ are open, then $(0,1)\times(0,1) \cup (2,3)\times(2,3)$ should also be open but I can't see why they can't be open. – 2017-01-31
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    @OliverG it is **not** the topology, it's only a base for the topology. – 2017-01-31
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    But why can't we say that it is the topology? I can't see how it violates the definition of a topology. – 2017-01-31
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    @OliverG Well, the fact that the collection is not closed under unions disqualifies it as a topology (the third axiom fails!) – 2017-01-31
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    How does it fail? Why exactly isn't $(0,1)\times(0,1) \cup (2,3)\times (2,3)$ open in $(R^2, T')$? – 2017-01-31
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    @OliverG It *is* open by definition but it is not in $\mathcal{B}$, so $\mathcal{B}$ is not a topology. – 2017-01-31
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    So since $B$ is the set of all $O_1\times O_2$ and there's no $O_1 \times O_2$ that makes that union, that implies that $B$ cannot be a topology? And since $B$ satisfies the definition of basis it is a basis for the topology? – 2017-01-31
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    @OliverG indeed, it satisfies the axioms for being the base for **a** topology and the generated topology (via unions) is by definition the product topology. – 2017-01-31
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    So (last question), a subset of a finite product $A_1 \times A_2\times...\times A_n$ is open iff it's of the form $O_1 \times O_2 \times ... \times O_n$ where each $O_i$ is open in its respective $X_i$? – 2017-01-31
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    @OliverG that *is* true, but requires proof. I gave a proof in the last month, search my answers – 2017-01-31
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    Wouldn't a proof be almost trivial? Since if $\prod A_i$ is open then it's a union of elements of the basis and is therefore each $A_i = \bigcup O_i$? Unless I'm missing something, of course. – 2017-01-31
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    @OliverG you can write an open square as a union of smaller open squares. The writing as a union is not unique – 2017-01-31
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    @OliverG You can also prove the lemma first that projections are open maps. – 2017-01-31
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    So what am I missing with: Since open sets are unions of basis elements, $\prod A_i = \bigcup \prod O_i$. Therefore $x_i \in A_i \Rightarrow x_i \in \bigcup O_i$ and therefore $x_i \in O_j$ for some arbitrary open set. Then since $\prod A_i = \bigcup \prod O_i$, if $x_i \in \bigcup O_i \Rightarrow x_i \in A_i$, and therefore $\Rightarrow x_i \in O_k \Rightarrow x_i \in A_i$? – 2017-01-31
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    @OliverG You use that the union of product set equals a product of the union set. Why would this hold in general? – 2017-02-01
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Clarification: Most general, the product topology is generated by the sub-base of $\pi_i^{-1}(\theta_i)$, where $\theta_i$ is a open set of the $i^{th}$ space and $\pi_i$ the canonical projection in the respective space. A sub-base is a family of open sets s.t. any set in base is a finite intersection of sets in the sub-base. So the sets in base are finite intersection of pre-image of canonical projection; it's hard to imagine but it's like fixing finite coordinates in a open set and leaving the rest free in the respective component of the product.

Said, that, a open set of the product is a union (of any cardinal) of this pseudo-boxes from the base. The base is like lego pieces and topology it's like all you can build with the base.

Extra-information: The point of having sub-bases is that, with them, you can build a topology from almost nothing. I mean, what you get is the smallest topology that contains the sub-base, which don't need to have any property. That's hoy we can give a topology to a so ware space like product (any cardinal)