TITLE: Proof That For Two Uncountable Sets, The Cardinality of the Collection of Their Respective Uncountable Subsets Are Equal.
QUESTION [1 upvotes]: The proposition I came along is as follows:
Assume $S$ & $T$ are uncountable sets. Then if $\mathcal{P}_{u}(S)$ and $\mathcal{P}_{u}(T)$ denotes the respective collection of uncountable subsets of $S$ & $T$,  $|\mathcal{P}_{u}(S)| = |\mathcal{P}_{u}(T)|$.
I did write out a proof on if $|S| = |T|$, then $|\mathcal{P}(S)| = |\mathcal{P}(T)|$ for the normal type of Power sets. This conclusion was derived from the fact that due to the existence of bijective function, $f: S \rightarrow T$, and $f^{-1}: T \rightarrow S$, all sets within each power set can be constructed by a mapping from a set $S_{i} \rightarrow T_{i}$ s.t. $S_{i} \subseteq S$ and $T_{i} \subseteq T$. However, the likely apparent connection between this proof and the former is still uncertain for myself.

REPLY [8 votes]: This is wildly not true. Take $T$ to be any uncountable set, you can show that $|\mathcal P(T)|=|\mathcal P_u(T)|$. Now take $S$ to be $\mathcal P(T)$ and disprove the claim. 
(Even without this additional claim, you can directly show this is false for $T$ any uncountable set and $S=\mathcal P(\mathcal P(T))$.)
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