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Thread: Properties of the product topology

  1. #1
    MHF Contributor Drexel28's Avatar
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    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 $\displaystyle \left\{X_j\right\}_{j\in\mathcal{J}}$ be a class of topological spaces and let $\displaystyle \mathfrak{B}_\ell$ be an open base for $\displaystyle X_\ell$ for each $\displaystyle \ell\in\mathcal{J}$. Then, $\displaystyle \mathfrak{B} = \prod_{j\in\mathcal{J}}\mathfrak{B}_j$ is an open base for $\displaystyle X = \prod_{j\in\mathcal{J}}X_j$ under the product topology.

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

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

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

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

    6. Consequently, if $\displaystyle X_1,\cdots,X_n$ are separable topological spaces then $\displaystyle X = X_1\times\cdots\times X_n$ 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.
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  2. #2
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    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 $\displaystyle \mathcal J$ being finite or not? If you pick an open neighbourhood $\displaystyle N$ of $\displaystyle \pi_j (x)$ in $\displaystyle X_j$, the inverse image of $\displaystyle N$ is open since $\displaystyle \pi_j$ is continuous, and it obviously contains $\displaystyle x$.
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  3. #3
    Member mabruka's Avatar
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    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.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    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 $\displaystyle \mathcal{J}$ 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 $\displaystyle \mathfrak{B}=\left\{B_1\times \cdots\times B_n:B_k\in\mathfrak{B}_k,\text{ }1\leqslant k\leqslant n\right\}$ 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 $\displaystyle \mathcal{J}$ 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 $\displaystyle \mathcal J$ being finite or not? If you pick an open neighbourhood $\displaystyle N$ of $\displaystyle \pi_j (x)$ in $\displaystyle X_j$, the inverse image of $\displaystyle N$ is open since $\displaystyle \pi_j$ is continuous, and it obviously contains $\displaystyle x$.
    This was just stupidity on my part. Of course you are correct, I wasn't thinking!


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  5. #5
    MHF Contributor Drexel28's Avatar
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    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 $\displaystyle \left\{X_j\right\}_{j\in\mathcal{J}}$ is a collection of topological spaces and $\displaystyle X = \prod_{j\in\mathcal{J}}X_j$ is the product space of these spaces, then for any $\displaystyle x\in X$ we have that if $\displaystyle N$ is a neighborhood of $\displaystyle X$ then $\displaystyle \pi_j(N)$ is a neighborhood of $\displaystyle \pi_j(x)$ for each $\displaystyle j\in\mathcal{J}$
    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 $\displaystyle \left\{X_j\right\}_{j\in\mathcal{J}}$ is a collection of topological spaces and $\displaystyle X = \prod_{j\in\mathcal{J}}X_j$ is the product space of these spaces, then for any $\displaystyle x\in X$ we have that if $\displaystyle N$ is a neighborhood of $\displaystyle X$ then $\displaystyle \pi_j(N)$ is a neighborhood of $\displaystyle \pi_j(x)$ for each $\displaystyle j\in\mathcal{J}$
    Proof: Let $\displaystyle N$ be a neighborhood of $\displaystyle x$ in $\displaystyle X$. Then, $\displaystyle N=\bigcup_{k\in\mathcal{K}}O_k$ where $\displaystyle O_k$ is an open basic set in the product topology. Then, $\displaystyle \pi_j\left(N\right)=\pi_j\left(\bigcup_{k\in\mathc al{K}}O_k\right)=\bigcup_{k\in\mathcal{K}}\pi_j\le ft(O_k\right)$. Now regardless of what $\displaystyle O_k$ we know that $\displaystyle \pi_j\left(O_k\right)$ is open (either it is the full space $\displaystyle X_j$ or some open set $\displaystyle G_j$) and so each $\displaystyle \pi_j\left(O_k\right)$ is an open set in $\displaystyle X_j$ and so $\displaystyle \pi_j\left(N\right)=\bigcup_{k\in\mathcal{K}}\pi_j \left(O_k\right)$ is the arbitrary union of open sets in $\displaystyle X_j$ and thus open in $\displaystyle X_j$. And of course lastly noting that $\displaystyle x\in N\implies \pi_j(x)\in\pi_j\left(N\right)$ finishes the argument.
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  6. #6
    Member mabruka's Avatar
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    I think 5) can be proved "directly".

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

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


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

    Then $\displaystyle x \in A \cap D$.


    so D is dense in X.
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  7. #7
    Member mabruka's Avatar
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    the proof of 3) you posted is good!

    you are right, thank you =)
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  8. #8
    MHF Contributor Drexel28's Avatar
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    In case anyone is interested here are my proofs:


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

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

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


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


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

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

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


    6. Consequently, if $\displaystyle X_1,\cdots,X_n$ are separable topological spaces then $\displaystyle X = X_1\times\cdots\times X_n$ is separable.
    Proof: Since $\displaystyle X_1,\cdots,X_n$ are separable there exists countable dense subsets $\displaystyle \mathcal{D}_1,\cdots,\mathcal{D}_n$ which are countable and since $\displaystyle \mathcal{D}_1\times\cdots\times\mathcal{D}_n$ is dense in $\displaystyle X_1\times\cdots\times X_n$ and the product of finitely many countable sets is countable the conclusion follows.
    Last edited by Drexel28; Feb 11th 2010 at 12:07 PM. Reason: LaTeX issue
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  9. #9
    Member mabruka's Avatar
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    how come 4) is not true?
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  10. #10
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mabruka View Post
    how come 4) is not true?
    Obvious counterexample, take the diagonal $\displaystyle \Delta\subset X\times X$ in any connected Hausdorff space
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  11. #11
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Drexel28 View Post

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

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

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

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

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

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

    Therefore, $\displaystyle X$ is second countable
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