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10h
comment What $+1+1+\cdots$ really equals
@Lovsovs: Specifically for $1+2+3+\ldots$, there is no integer $p$ where the sum makes sense in the $p$-adic numbers.
12h
comment Does the following condition implies full outer measure?
Other than the user who posted this answer, is there anything actually wrong with it?
12h
revised Cardinality of infinite sets
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12h
comment Cardinality of infinite sets
See also: math.stackexchange.com/q/182236
12h
comment Why is the Axiom of Infinity necessary?
In addition to what @Noah said, what does WM has to do with this question? The quote is actually correct. $\sf PA$ is not adequate for establishing the existence of any set, since sets do not exist in models of $\sf PA$. But even if you talk about theories which are bi-interpretable with $\sf PA$, you still don't get the axiom of infinity which is necessary for proving the existence of an infinite set, e.g. the set of all natural numbers.
12h
comment Cardinality of infinite sets
The theorem clearly cannot be extended this way because of the many counterexamples that existed before Cantor was even born.
12h
revised uncountable well-ordered chain in $(\mathcal{P}(\mathbb{N}),\subseteq)$ without $AC$
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12h
comment uncountable well-ordered chain in $(\mathcal{P}(\mathbb{N}),\subseteq)$ without $AC$
@Noah: You're right about the first part. I'm not sure about the second. If you map $\alpha$ to the least $n$ which wasn't used thus far, and does not appear in any previous set, you should get an injection into $\omega$. You're right about successors being the easy way out, though. Map $\alpha$ to the least element in $A_{\alpha+1}\setminus A_\alpha$, and you definitely get an injection into $\omega$.
13h
comment uncountable well-ordered chain in $(\mathcal{P}(\mathbb{N}),\subseteq)$ without $AC$
@Alessandro: Yes, thanks!
13h
revised uncountable well-ordered chain in $(\mathcal{P}(\mathbb{N}),\subseteq)$ without $AC$
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13h
revised uncountable well-ordered chain in $(\mathcal{P}(\mathbb{N}),\subseteq)$ without $AC$
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13h
comment uncountable well-ordered chain in $(\mathcal{P}(\mathbb{N}),\subseteq)$ without $AC$
I'm fairly sure that it also works for the chain $\{\{0,\ldots,n-1\}\mid n\in\Bbb N\}$ which has order type $\omega$. :-) Are you claiming $\omega$ is a successor ordinal? :-P
13h
answered uncountable well-ordered chain in $(\mathcal{P}(\mathbb{N}),\subseteq)$ without $AC$
13h
revised Proof on Functions /Set Theory
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13h
revised Proving the existence of a proof without actually giving a proof
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16h
comment Russell's paradox question
Have you read the question?
16h
revised proof of 1) $|a| = 0 \iff a = 0$ and 2) $|a| \ge 0$
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17h
comment Russell's paradox question
Can you be a bit more specific?
17h
revised Russell's paradox question
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17h
answered Russell's paradox question