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 Yearling
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Jan
31
accepted Flipping an unfair coin n times
Jan
31
asked Flipping an unfair coin n times
Dec
4
awarded  Yearling
Jul
2
awarded  Curious
Dec
30
accepted Axiom of Choice and Recursion Theorem exercise
Dec
30
revised Axiom of Choice and Recursion Theorem exercise
deleted 25 characters in body
Dec
30
asked Axiom of Choice and Recursion Theorem exercise
Dec
30
comment A property of functions from $\mathbb{R}\rightarrow \mathbb{R}$
@DanielFischer, I may assume the continuum hypothesis. Thank you for helping out.
Dec
30
comment A property of functions from $\mathbb{R}\rightarrow \mathbb{R}$
@DanielFischer The countable union of the sets $M_k$ must add up to the domain of $f$ which is $\mathbb{R}$. If all $M_k$ are countable then their union ia also countable, a contradiction. Thus at least one has the cardinality of $\mathbb{R}$. Did I got it right?
Dec
30
asked A property of functions from $\mathbb{R}\rightarrow \mathbb{R}$
Dec
24
comment Is my claim wrong?
@AndreasBlass you're right of course... I really liked my proof lol I'll try and prove your claim (more homework lol) Don't be too surprised if you see me posting again in few days:) . You've been most helpfull, thank you again.
Dec
24
comment Is my claim wrong?
@AndreasBlass After a long night I think I managed to prove the following (which actually proves the claim about the existance of a monomorphism): If $U,V$ are well ordered sets and $\pi:U\rightarrow V$ a monomorphism which respects the ordering ($x\leq_U y\iff \pi(x)\leq_V \pi(y)$) then $p[U]=seg_V(w)=\{y\in V| y<w\}$. So yes you're right there is a $a_0\in\chi(\mathbb{N})$ such that for all $a>a_0$ I don't have $M_a$ for most $a\in\chi(\mathbb{N})$. Andreas thank you for your time. If you post your comment as an answer I'll accept it.
Dec
23
comment Hartogs space of $\mathbb{N}$
@CameronBuie $A\leq_0 B$ means there exists a monomorphism $\pi$ from $A$ to $B$ which respects the ordering $(x\leq_{A} y\iff \pi(x)\leq_{B} \pi(y))$
Dec
23
answered Order of elements within a group
Dec
23
asked Is my claim wrong?
Dec
23
accepted Hartogs space of $\mathbb{N}$
Dec
23
comment Hartogs space of $\mathbb{N}$
@AsafKaragila thank you for helping out, I'll work a bit more on the details.
Dec
23
comment Hartogs space of $\mathbb{N}$
@AsafKaragila yes but I'm still working on a few things. Following your hint I did the following (I'm not really sure though about my solution): Since $\chi(\mathbb{N})$ is well ordered it has a least element let's call it $0$ and then $M_0$ is also the first set in the sequence defined above. Assuming the set sequence doesn't become 'constant' I can find $b,c\in \chi(\mathbb{N})$ with $b<c$ such that $M_c\subset M_b$. I can pick $x_0\in\M_0$ such that $x_0\in M_b\setminus M_c$. I then move to the next set etc.Picking different elements from each set I have a contradiction $M_0>_c \mathbb{N}$
Dec
23
revised Hartogs space of $\mathbb{N}$
deleted 517 characters in body
Dec
22
revised Hartogs space of $\mathbb{N}$
added 17 characters in body