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I ran upon this topology based on binary space, perhaps using obscure terminology, but I am curious what it is and its properties.

Let binary space be the set of strings of $0,1$'s, and let $S$ be the set of all functions that map the binary space to a set of two elements, $\{0,1\}$. It's like it decides a true or false for every binary string. For the topology $\mathcal{T}$, let a basic set be $U_{V,f}$, where any $g$ in this set must satisfy $g(x)=f(x)$ for all $x$ in $V$, and $V$ is a finite subset of $B$.

Is this space $(S,\mathcal{T})$ Hausdorff? Compact? Any related material is welcome.

Thanks

verticese
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  • Do they specify that $V$ is finite? – David C. Ullrich Apr 17 '16 at 16:23
  • Sorry forgot to add that $V$ is finite. – verticese Apr 17 '16 at 16:24
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    Then this topology is just the product topology on ${0,1}^{\Bbb N}$; so it's compact and hausdorff. – David C. Ullrich Apr 17 '16 at 16:26
  • Do we have both finite strings and infinite ones, or just the finite ones, or just the infinite ones? string is ambiguous. – Henno Brandsma Apr 17 '16 at 16:48
  • I think both are allowed in the binary space. – verticese Apr 17 '16 at 16:57
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    Here it is not: http://math.stackexchange.com/q/972654/4280, just the infinite ones. – Henno Brandsma Apr 17 '16 at 18:11
  • @DavidC.Ullrich Could you please explain why it is the product topology? The space contains functions that maps {0,1}^N to {0,1}, but how can you relate it to the product topology? Thanks – verticese Apr 18 '16 at 16:13
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    I misread the question, but it doesn't matter. If $B$ is "binary space" then $S$ is exactly ${0,1}^B$, and the topology you describe is the product topology. (Do you know the definition of $A^B$ and the definition of the product topology? I'm having a hard time seeing where the problem is...) – David C. Ullrich Apr 18 '16 at 19:03
  • @DavidC.Ullrich Could you please point me to the definition of $A^B$? I know product topology but have a hard time seeing the connection. – verticese Apr 19 '16 at 15:27
  • I don't understand how what you say is possible. If you know what a product topology is you must know what a product is, namely a certain space of functions. Whatever. By definition $A^B$ is the set of all functions from $B$ to $A$. To put it in other words, $A^B=\prod_{b\in B}A$. Or in notation that may make more sense or less sense, $A^B=\prod_{b\in B}A_b$, where $A_b=A$ for all $b$. What's the definition of the product topology again? – David C. Ullrich Apr 19 '16 at 15:39
  • @DavidC.Ullrich Its elements are of the from $\prod x_i$, and open sets are the union of the product of the open sets in the original topologies? – verticese Apr 19 '16 at 15:45
  • No. Not quite that simple. Whoever assigned this exercise intended it as an exercise in the product topology and the Tychonoff theorem, etc. You need to get the definitions straight before attempting the problem. – David C. Ullrich Apr 19 '16 at 15:47

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The set of finite binary sequences $B$ is denoted $\{0,1\}^{<\omega}$. Your set $S$ is the set of functions $f:\{0,1\}^{<\omega}\to \{0,1\}$. That is, $$S=\{0,1\}^{\{0,1\}^{<\omega}}.$$

You defined a basic open set around $f\in S$ to be the set of all $g\in S$ which agree with $f$ on a given finite subset of ${\{0,1\}^{<\omega}}$. This is exactly how basic open sets are defined in the product topology. As the product of compact Hausdorff spaces (in your case $\{0,1\}$) is compact Hausdorff in the product topology, $S$ is compact Hausdorff. Do not get hung up on the index set $\{0,1\}^{<\omega}$. You could replace it with any other countably infinite set (such as $\omega$) and get the same space.

Forever Mozart
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