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Does anyone know a non trivial(that we cannot define on every set) topology defined on the power set of an uncountable set?

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What does "trivial" mean here? The cofinite, cocountable, and particular point topologies (to name just the first three that come to mind) can be defined on any set. Are they trivial? –  Chris Eagle Jan 8 '11 at 7:59
    
I corrected it.By trivial here i mean the topologies that can be defined on any set. –  t.k Jan 8 '11 at 8:05
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3 Answers 3

up vote 7 down vote accepted

I think you're asking whether power objects exist in the category of topological spaces. I think the answer is not always: although we have a subobject classifier $\Omega$ (namely the 2-point space with the indiscrete topology), and the exponential object $\Omega^X$ is not guaranteed to exist. The obstruction is the requirement that for every continuous function $f: A \times X \to \Omega$ there must be a unique continuous function $\tilde{f}: A \to \Omega^X$, such that $\tilde{f}(a)(x) = f(a, x)$, and vice-versa. When $X$ is nice enough, e.g. locally compact and Hausdorff, then $\Omega^X$ does exist and has the compact-open topology.

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That's exactly what i wanted to find out! –  t.k Jan 8 '11 at 10:36
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You may find this helpful: "A Topology on a Power Set of a Set and Convergence of a Sequence of Sets."

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Current link is here. (I would have edited it into the answer but the system won't allow me to change just one character.) –  Ari Brodsky May 24 '13 at 6:40
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I don't quite follow your question. But I provide one example anyway. $2^{[0,1]}$ is an example which is compact but not sequentially compact. In fact, we choose a series of points $f_n\in 2^{[0,1]}$ as follows. We definte $f_n(x)$ as the nth digit after writing x in the binary system.

Then, when we choose any subsequence $f_{n_j}$ of $f_n$, we can find a $x\in [0,1]$ such that, after rewriting it in the binary system, it has $n_1$th,$n_2$th, $\cdots$ digits $0,1,0,1,\cdots$, alternatively. Obviously, $f_{n_j}$ has its value in x $0,1,0\cdots$ and thus doesn't have a limit.

But by Tychonoff's theorem, we know this space is compact.

updated: $2^{[0,1]}$ is endowed with product topology.

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I am asking what a topological "open" set would have to be like in the power set –  t.k Jan 8 '11 at 8:20
    
@t.spero: Xiaochuan is using the product topology on $2^{[0,1]}$. (The reference to Tycohoff's theorem is a clue!) More information is available on wikipedia. In concrete terms, if your base set is $S$, open sets on $\mathcal{P}(S)$ are generated by the sets $\mathcal{U}(F, G)$ for finite sets $F$ and $G \in S$, where $\mathcal{U}(F,G)$ is defined to be $\{U \subset S : F \subset U \text{ and } G \cap U = \emptyset\}$. (I believe I have that right, but I am also tired, so no guarantees.) –  Jon Jan 8 '11 at 9:32
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