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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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    I corrected it.By trivial here i mean the topologies that can be defined on any set.2011-01-08

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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 exactl$y$ what i wanted to find out!2011-01-08
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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](http://www1.au.edu.tw/ox_view/edu/tojms/j_paper/Full_text/Vol-21/No-1/21%281%29-2-0333.pdf). (I would have edited it into the answer but the system won't allow me to change just one character.)2013-05-24
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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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    @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](http://en.wikipedia.org/wiki/Product_topology). 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.)2011-01-08