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Thread: Show that X is compact

  1. #1
    MHF Contributor alexmahone's Avatar
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    Show that X is compact

    Consider the two-element set $\{0,\ 1\}$ equipped with the discrete topology, and form the countably infinite product

    $\displaystyle X:=\{0,\ 1\}^\omega=\prod\limits_{n\in\mathbb{Z_+}}\{0,\ 1\}$

    So $X$ consists of the infinite sequences $\displaystyle(x_n)_{n\in\mathbb{Z_+}}$, where for each $k\in\mathbb{Z}_+$, the $k$th term $x_k$ is either $0$ or $1$. Equip $X$ with the product topology.

    Show that $X$ is compact (you may not use Tychonoff’s theorem).
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    Re: Show that X is compact

    Quote Originally Posted by alexmahone View Post
    Consider the two-element set $\{0,\ 1\}$ equipped with the discrete topology, and form the countably infinite product

    $\displaystyle X:=\{0,\ 1\}^\omega=\prod\limits_{n\in\mathbb{Z_+}}\{0,\ 1\}$

    So $X$ consists of the infinite sequences $\displaystyle(x_n)_{n\in\mathbb{Z_+}}$, where for each $k\in\mathbb{Z}_+$, the $k$th term $x_k$ is either $0$ or $1$. Equip $X$ with the product topology.

    Show that $X$ is compact (you may not use Tychonoff’s theorem).
    Start with definitions. Define the product topology. What open sets are there? Next, use the definition of compact. The Cantor Intersection Theorem may be more useful than the fact that every open cover of X has a finite subcover, but it has been some time since I did this particular problem, so both definitions for compactness may apply. Once you list the definitions out here in this post, see what inferences you can make. Once you get stuck, then we will help you further.
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    MHF Contributor alexmahone's Avatar
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    Re: Show that X is compact

    Quote Originally Posted by SlipEternal View Post
    Start with definitions. Define the product topology. What open sets are there? Next, use the definition of compact. The Cantor Intersection Theorem may be more useful than the fact that every open cover of X has a finite subcover, but it has been some time since I did this particular problem, so both definitions for compactness may apply. Once you list the definitions out here in this post, see what inferences you can make. Once you get stuck, then we will help you further.
    I haven't learnt the Cantor Intersection Theorem.

    Let $\mathcal{A}=\{A_\alpha\}$ be an open cover of $X$.

    $\implies A_\alpha$'s are open sets of $X$ and $\bigcup\limits_\alpha A_\alpha=X$.

    The problem is that it's not easy to describe the $A_\alpha$'s. The basis elements of $X$ can be easily described: they are infinite cartesian products whose components are subsets of $\{0, 1\}$.
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    Re: Show that X is compact

    Quote Originally Posted by alexmahone View Post
    I haven't learnt the Cantor Intersection Theorem.

    Let $\mathcal{A}=\{A_\alpha\}$ be an open cover of $X$.

    $\implies A_\alpha$'s are open sets of $X$ and $\bigcup\limits_\alpha A_\alpha=X$.

    The problem is that it's not easy to describe the $A_\alpha$'s. The basis elements of $X$ can be easily described: they are infinite cartesian products whose components are subsets of $\{0, 1\}$.
    Again, what is the product topology? That is what is important to define here. A set is open in the product topology if and only if it contains a subset of $\{0,1\}$ in only a finite number of coordinates. Let $I = \{0,1\}$. Then you have $I\times I \times I \times \{0\} \times I \times I \times \cdots \times I \times \{1\} \times I \times \cdots$ and you have only a finite number of coordinates in the product that are not $I$. So, you can represent an open set by the coordinates where it is not the full set $\{0,1\}$.

    Let's take any open cover $\mathcal{O}$ of $X$. Take any open set $O \in \mathcal{O}$ in that open cover. It will be $I$ in every coordinate except a finite number of coordinates. Let's say it is coordinates $i_1, \ldots, i_n$. In those coordinates, you have either $\{0\}$ or $\{1\}$.

    You want to find other open sets that must be in $\mathcal{O}$ such that you close up any holes you may have. Do you see a strategy for doing this? If not, get as far as you can and I can provide more hints.
    Last edited by SlipEternal; Apr 23rd 2018 at 10:03 AM.
    Thanks from topsquark
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    MHF Contributor alexmahone's Avatar
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    Re: Show that X is compact

    Quote Originally Posted by SlipEternal View Post
    A set is open in the product topology if and only if it contains a subset of $\{0,1\}$ in only a finite number of coordinates. Let $I = \{0,1\}$. Then you have $I\times I \times I \times \{0\} \times I \times I \times \cdots \times I \times \{1\} \times I \times \cdots$ and you have only a finite number of coordinates in the product that are not $I$. So, you can represent an open set by the coordinates where it is not the full set $\{0,1\}$.
    I think you're describing the basis elements of the product topology, not the open sets.
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    Re: Show that X is compact

    Quote Originally Posted by alexmahone View Post
    I think you're describing the basis elements of the product topology, not the open sets.
    No, I am describing the entire product topology. The basis elements would be the sets where the infinite product is a subset of $I$ in exactly 1 coordinate rather than in a finite number of coordinates. Suppose you take all unions and finite intersections of these basis elements. You will get a set that is not the full $I$ at only finitely many coordinates. If you still have questions about the product topology, please describe what you think an open set would look like.
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