Topology of the power set Does anyone know a non trivial(that we cannot define on every set) topology defined on the power set of an uncountable set?
 A: You may find this helpful: "A Topology on a Power Set of a Set and Convergence of a Sequence of Sets."
A: 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.
A: 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.
