Results 1 to 11 of 11

Math Help - Properties of the product topology

  1. #1
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21

    Properties of the product topology

    Hello friends. I am doing an independent study course, and it is a bit of the Moore method style. So, right now I am studying product topology and have come up with some conjectures. I have "proof" for all of them but would appreciate (no need for proof if you don't want) if someone could validate whether or not they are true.

    Conjectures:

    1. Let \left\{X_j\right\}_{j\in\mathcal{J}} be a class of topological spaces and let \mathfrak{B}_\ell be an open base for X_\ell for each \ell\in\mathcal{J}. Then, \mathfrak{B} = \prod_{j\in\mathcal{J}}\mathfrak{B}_j is an open base for X = \prod_{j\in\mathcal{J}}X_j under the product topology.

    2. Consequently, if X_1,\cdots,X_n are a finite collection of second countable topological spaces then X_1\times\cdots\times X_n is separable.

    3. If \left\{X_j\right\}_{j\in\mathcal{J}} is a collection of topological spaces and X = \prod_{j\in\mathcal{J}}X_j is the product space of these spaces, then for any x\in X we have that if N is a neighborhood of X then \pi_j(N) is a neighborhood of \pi_j(x) for each j\in\mathcal{J}

    4. The converse is true if \mathcal{J} is finite.

    5. Using this we can show that if \left\{X_j\right\}_{j\in\mathcal{J}} is a collection of topological spaces, and \mathcal{D}_\ell is dense in X_\ell for each \ell\in \mathcal{J} then \prod_{j\in\mathcal{J}}\mathcal{D}_j is dense in X = \prod_{j\in\mathcal{J}}X_j with the product topology.

    6. Consequently, if X_1,\cdots,X_n are separable topological spaces then X = X_1\times\cdots\times X_n is separable.

    That's it for now.

    Any input would be incredibly appreciated. Also, I feel as though I should point out that even though I said this is Moore method like...this is just for my own learning. There is no attempt at foul play here.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Feb 2010
    From
    Lisbon
    Posts
    51
    Well, here are 3 I'm pretty sure of. I'll see if I manage to look at the rest, but I need more time

    1. False. First off, the product cannot be written like you did. You're taking the product of the elements of the bases, not of the bases themselves. That aside, you need to restrict the product to only finitely many indices (the rest being the whole space).

    2. A countable (need not be finite) product of second countable spaces is second countable, and every second countable space is separable.

    4. Isn't it always true, independently of \mathcal J being finite or not? If you pick an open neighbourhood N of \pi_j (x) in X_j, the inverse image of N is open since \pi_j is continuous, and it obviously contains x.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member mabruka's Avatar
    Joined
    Jan 2010
    From
    Mexico City
    Posts
    150
    3) Is false unless something else is suposed.

    As Nyrix said continuity of projections guarantee the validity of 4) in ANY case.

    Maybe 3 is valid only if J is finite ?

    IF the product is finite the product topology and "box"-topology are the same, under the box topology i think 3) is true by construction.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Nyrox View Post
    Well, here are 3 I'm pretty sure of. I'll see if I manage to look at the rest, but I need more time

    1. False. First off, the product cannot be written like you did. You're taking the product of the elements of the bases, not of the bases themselves. That aside, you need to restrict the product to only finitely many indices (the rest being the whole space).
    My mistake, I should have said that \mathcal{J} is finite. Otherwise, as you pointed out there is no reason for the base elements to even be open under the product topology. Also, I had a bit of a notational faux pas. What I meant by the product was this. The set \mathfrak{B}=\left\{B_1\times \cdots\times B_n:B_k\in\mathfrak{B}_k,\text{ }1\leqslant k\leqslant n\right\} is an open base.

    2. A countable (need not be finite) product of second countable spaces is second countable, and every second countable space is separable.
    I see, but from what I have proved in the above (with the addendum that \mathcal{J} is finite) it follows relatively easily that the finite case is true. I will look more closely at the countable case.

    4. Isn't it always true, independently of \mathcal J being finite or not? If you pick an open neighbourhood N of \pi_j (x) in X_j, the inverse image of N is open since \pi_j is continuous, and it obviously contains x.
    This was just stupidity on my part. Of course you are correct, I wasn't thinking!


    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by mabruka View Post
    3) Is false unless something else is suposed.

    As Nyrix said continuity of projections guarantee the validity of 4) in ANY case.

    Maybe 3 is valid only if J is finite ?

    IF the product is finite the product topology and "box"-topology are the same, under the box topology i think 3) is true by construction.
    Quote Originally Posted by Drexel28 View Post

    3. If \left\{X_j\right\}_{j\in\mathcal{J}} is a collection of topological spaces and X = \prod_{j\in\mathcal{J}}X_j is the product space of these spaces, then for any x\in X we have that if N is a neighborhood of X then \pi_j(N) is a neighborhood of \pi_j(x) for each j\in\mathcal{J}
    It is actually true in general. I had it validated by someone who is algebraic topologist on AOPS!

    Here is the proof

    Quote Originally Posted by Drexel28 View Post

    3. If \left\{X_j\right\}_{j\in\mathcal{J}} is a collection of topological spaces and X = \prod_{j\in\mathcal{J}}X_j is the product space of these spaces, then for any x\in X we have that if N is a neighborhood of X then \pi_j(N) is a neighborhood of \pi_j(x) for each j\in\mathcal{J}
    Proof: Let N be a neighborhood of x in X. Then, N=\bigcup_{k\in\mathcal{K}}O_k where O_k is an open basic set in the product topology. Then, \pi_j\left(N\right)=\pi_j\left(\bigcup_{k\in\mathc  al{K}}O_k\right)=\bigcup_{k\in\mathcal{K}}\pi_j\le  ft(O_k\right). Now regardless of what O_k we know that \pi_j\left(O_k\right) is open (either it is the full space X_j or some open set G_j) and so each \pi_j\left(O_k\right) is an open set in X_j and so \pi_j\left(N\right)=\bigcup_{k\in\mathcal{K}}\pi_j  \left(O_k\right) is the arbitrary union of open sets in X_j and thus open in X_j. And of course lastly noting that x\in N\implies \pi_j(x)\in\pi_j\left(N\right) finishes the argument.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member mabruka's Avatar
    Joined
    Jan 2010
    From
    Mexico City
    Posts
    150
    I think 5) can be proved "directly".

    Let A be an open set in  X, so A is a finite intersection of open sets, say  A_i_1,...,A_i_n in  X_i_1,...,X_i_n respectively.

    Every D_j is dense so theres an d_{i_j} \in A_{i_j} \cap D_{i_j} for each j=1,...,n


    Let x \in X the element which \pi_{i_j} (x)=d_{i_j} for j=1,...,n and \pi_i(x)= a_i where p_i is any element (AXIOS OF CHOICE) of D_i.<br />

    Then x \in A \cap D.


    so D is dense in X.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member mabruka's Avatar
    Joined
    Jan 2010
    From
    Mexico City
    Posts
    150
    the proof of 3) you posted is good!

    you are right, thank you =)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    In case anyone is interested here are my proofs:


    Conjectures:
    1. Let \left\{X_j\right\}_{j\in\mathcal{J}} be a class of topological spaces and let \mathfrak{B}_\ell be an open base for X_\ell for each \ell\in\mathcal{J}. Then, \mathfrak{B} = \left\{B_1\times\cdots\times B_n:B_k\in\mathfrak{B}_k,\text{ }1\leqslant k\leqslant n\right\} is an open base for X = \prod_{j\in\mathcal{J}}X_j under the product topology. If \mathcal{J} is finite
    Proof: It is clear that since the box topology coincides with the product topology for the product of a finite number of spaces that the elements of \mathfrak{B} are open, so it just remains to show that they form a base.

    So, let O be open in X. Since the sets of the form G=G_1\times\cdots\times G_n are an open base with G_k open in X_k it follows that for any o\in O there exists some G such that o\in G\subseteq O. But considering 3. (already proved) we see that \pi_\ell(G),\text{ }1\leqslant \ell\leqslant n is an open set containing \pi_\ell(o) and so there exists some E_\ell\in\mathfrak{B}_\ell such that \pi_\ell(o)\in E_\ell\subseteq\pi_\ell(G). Clearly then since each E_1,\cdots,E_n is open we have that E_1\times\cdots\times E_n is an open set which is a subset of G_1\times\cdots\times G_n and which contains o. The conclusion follows.

    2. Consequently, if X_1,\cdots,X_n are a finite collection of second countable topological spaces then X_1\times\cdots\times X_n is second countable.
    Proof: Since X_1,\cdots,X_n are second countable there exists countable bases \mathfrak{B}_1,\cdots,\mathfrak{B}_n for each. As stated though, the set \mathfrak{B}=\left\{B_1\times\cdots\times B_n:B_k\in\mathfrak{B}_k,\text{ }1\leqslant k\leqslant n\right\} is an open base for X. So lastly noting that the mapping \eta:\prod_{j=1}^{n}\mathfrak{B}_j\mapsto\mathfrak  {B} given by \left(B_1,\cdots,B_n\right)\mapsto B_1\times\cdots\times B_n is a bijection finishes the argument since the \prod_{j=1}^{n}\mathfrak{B}_j is the finite product of countable sets, and thus countable.


    3. If \left\{X_j\right\}_{j\in\mathcal{J}} is a collection of topological spaces and X = \prod_{j\in\mathcal{J}}X_j is the product space of these spaces, then for any x\in X we have that if N is a neighborhood of X then \pi_j(N) is a neighborhood of \pi_j(x) for each j\in\mathcal{J}
    Already proved.


    4. The converse is true if \mathcal{J} is finite.
    NOT TRUE!

    5. Using this we can show that if \left\{X_j\right\}_{j\in\mathcal{J}} is a collection of topological spaces, and \mathcal{D}_\ell is dense in X_\ell for each \ell\in \mathcal{J} then \prod_{j\in\mathcal{J}}\mathcal{D}_j is dense in X = \prod_{j\in\mathcal{J}}X_j with the product topology.
    Proof: Let x\in X be arbitrary and let N be a neighborhood of x. By 3. we know that \pi_j\left(N\right) is a neighborhood of \pi_j(x) for each j\in\mathcal{J} and since \mathfrak{D}_j is dense in X_j we know there must exists some d_j\in\mathfrak{D}_j such that d_j\in\pi_j\left(N\right). Doing this for each j\in\mathcal{J} we see that \prod_{j\in\mathcal{J}}\{d_j\}\in\prod_{j\in\mathc  al{J}}\mathcal{D}_j is in N. The conclusion follows.

    Alternatively, I believe it is correct that \prod_{j\in\mathcal{J}}\overline{E_j}=\overline{\p  rod_{j\in\mathcal{J}}E_j} from where it would follow directly since \overline{\prod_{j\in\mathcal{J}}\mathcal{D}_j}=\p  rod_{j\in\mathcal{J}}\overline{\mathcal{D}_j}=\pro  d_{j\in\mathcal{J}}X_j=X


    6. Consequently, if X_1,\cdots,X_n are separable topological spaces then X = X_1\times\cdots\times X_n is separable.
    Proof: Since X_1,\cdots,X_n are separable there exists countable dense subsets \mathcal{D}_1,\cdots,\mathcal{D}_n which are countable and since \mathcal{D}_1\times\cdots\times\mathcal{D}_n is dense in X_1\times\cdots\times X_n and the product of finitely many countable sets is countable the conclusion follows.
    Last edited by Drexel28; February 11th 2010 at 01:07 PM. Reason: LaTeX issue
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member mabruka's Avatar
    Joined
    Jan 2010
    From
    Mexico City
    Posts
    150
    how come 4) is not true?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by mabruka View Post
    how come 4) is not true?
    Obvious counterexample, take the diagonal \Delta\subset X\times X in any connected Hausdorff space
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Drexel28 View Post

    2. Consequently, if X_1,\cdots,X_n are a finite collection of second countable topological spaces then X_1\times\cdots\times X_n is separable.
    I think you can extend this to say

    If \left\{X_n\right\}_{n\in\mathbb{N}} is a countable collection of second countable spaces, then \prod_{j=1}^{\infty}X_j is second countable
    Proof: Let \mathfrak{B}_\ell be corresponding countable base for X_\ell and define \mathcal{B}_n=\left\{E:\pi_{k}(E)\in\mathfrak{B}_k  ,\text{ }1\leqslant k\leqslant n,\text{ and }\pi_k(E)=X_k,\text{ }k>n\right\}. It is cleary by definition that each element of \mathcal{B}_n is open in X

    Next, we let \eta:\mathcal{B}_n\mapsto\mathfrak{B}_1\times\cdot  s\times\mathfrak{B}_n by \left(B_1,\cdots,B_n\right)\mapsto B_1\times\cdots\times B_n. It is clear that \eta is a bijection, and since \mathfrak{B}_1\times\cdots\times\mathfrak{B}_n is the finite product of countable sets we see that \mathcal{B}_n is countable.

    So, let \mathfrak{B}=\bigcup_{j=1}^{\infty}\mathcal{B}_j and let O\in X be an arbitrary open set. Since the set \mathcal{S} which consists of all sets E\subseteq X such that \pi_{k}(E) is open for all k and equals X_k for all but finitely many k is the defining open base for X we know that given any o\in O there exists some S\in \mathcal{S} such that o\in S\subseteq O. Let k_1,\cdots,k_n be the set of values such that \pi_{k_\ell}(S)\ne X_{k_{\ell}} and let m=\max\{k_1,\cdots,k_n\}

    Since \pi_{r}(S) is a neighborhood of o for 1\leqslant r\leqslant m and \mathfrak{B}_r is an open base for X_r there exists some B_r\in\mathfrak{B}_r such that \pi_r(o)\in B_r\subseteq\pi_r(S). So, let B\subseteq X be such that B=\prod_{z=1}^{\infty}G_z where G_z=B_z,\text{ }1\leqslant z\leqslant m and G_z=X_z,\text{ }z>m. Clearly o\in B\subseteq S\subseteq O and B\in\mathfrak{B}.

    It follows that \mathfrak{B} is an open base for X and since it's the countable union of countable sets it's countable.

    Therefore, X is second countable
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Product topology
    Posted in the Differential Geometry Forum
    Replies: 8
    Last Post: March 4th 2009, 12:35 PM
  2. Product topology
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: March 1st 2009, 06:04 AM
  3. discrete topology, product topology
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: December 13th 2008, 03:19 PM
  4. discrete topology, product topology
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: December 7th 2008, 02:01 PM
  5. Inner Product Properties
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: October 19th 2008, 02:09 PM

Search Tags


/mathhelpforum @mathhelpforum