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?

  • 1
    $\begingroup$ 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 $\endgroup$ – Julio Maldonado Henríquez Jan 31 '17 at 18:28

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).

  • $\begingroup$ 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. $\endgroup$ – Oliver G Jan 31 '17 at 18:49
  • $\begingroup$ @OliverG it is not the topology, it's only a base for the topology. $\endgroup$ – Henno Brandsma Jan 31 '17 at 18:50
  • $\begingroup$ But why can't we say that it is the topology? I can't see how it violates the definition of a topology. $\endgroup$ – Oliver G Jan 31 '17 at 18:52
  • $\begingroup$ @OliverG Well, the fact that the collection is not closed under unions disqualifies it as a topology (the third axiom fails!) $\endgroup$ – Henno Brandsma Jan 31 '17 at 18:53
  • $\begingroup$ How does it fail? Why exactly isn't $(0,1)\times(0,1) \cup (2,3)\times (2,3)$ open in $(R^2, T')$? $\endgroup$ – Oliver G Jan 31 '17 at 18:59

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)


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.