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 Dec 4 awarded Yearling 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 $p[U]=seg_V(w)=\{y\in V| ya_0$ 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 $b_c \mathbb{N}$ Dec 23 revised Hartogs space of $\mathbb{N}$ deleted 517 characters in body