2 votes

I am missing some point about Cantor's Theorem

You are forgetting the fact that $f$ is surjective onto the power set of $A$. If $A$ is a singleton then power set of $A$ is not a sigleton. It contains the empty set also. Existence of $\xi$ in the ...
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  • 2,176
1 vote
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Equality with metric in metric spaces

You first part of the $\impliedby$ proof has a mistake. You write "suppose $x < y$" but you should have written "suppose $x \ne y$". You've also been a little slopping elsewhere,...
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1 vote
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The map $g \mapsto \phi_g$ forms an isometry from $L^q$ to $(L^p)^*$

Your second sentence is wrong and probably shows the reason for your confusion: It is the map which sends $g$ to $\varphi_g$, that is $J(g):=\varphi_g$, which constitutes the isometry $J\colon L^q\to(...
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  • 1,543
1 vote
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Isomorphism between $L^q$ and $(L^p)^*$

If $g_1,g_2 \in L^{q}$ and $\int fg_1=\int fg_2$ for all $f \in L^{p}$ then $\int_E g_1 =\int_E g_2$ for every set $E$ of finite measure. This implies that $g_1=g_2$ almost everywhere.
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  • 2,176
1 vote
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Prove this limit in the weak topology of $L^2([0,T])$

For the sake of simplicity, assume that $T = 1$. We can prove that the result holds even if $f \in L^1$, i.e., \begin{equation*} \forall f \in L^1([0,1]), \qquad \int_0^1 \sum_{i=0}^{n-1} \frac{t_{i+1}...
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  • 4,854
1 vote

Given gog(x) Find g(x)

Hint: Let $g(x) = ax + b \implies (g \circ g)(x) = a(g(x)) + b$
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  • 61
1 vote

Hilbert space H is separable if and only if the dual space H* is separable

The Riesz representation theorem implements a conjugate-linear isometry $$\Phi: H \to H^*: \xi \mapsto \langle \xi\mid -\rangle$$(assuming that inner product is linear in the second factor). In ...
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