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Dec
30
comment LIM is cofinal in ON
I think I have figured it out now: Assume $M$ is a cofinal subset of ON. Then for every $\alpha$ in ON there is a $\beta$ in $M$ such that $\alpha \leq \beta$. Then consider $\bigcup M + 1$. This is an ordinal. But there is no $m \in M$ such that $\bigcup M + 1 \leq m$, therefore $M$ cannot be a set.
Dec
30
comment LIM is cofinal in ON
Where do you use cofinality in the first part of your proof?
Dec
30
comment LIM is cofinal in ON
I was quite confused yesterday. Now I think that in your first comment the second part is superfluous. To show that a cofinal subclass is proper it's enough to assume that it's a set and produce a contradiction.
Dec
30
comment LIM is cofinal in ON
I thought the von Neumann universe was the universe of all sets. So it's bigger than ON.
Dec
30
comment LIM is cofinal in ON
Can this proof be modified into not using the von Neumann universe?
Dec
29
comment LIM is cofinal in ON
Thank you, Asaf.
Dec
29
comment LIM is cofinal in ON
Is showing that every cofinal class of ON is proper also a one liner?
Dec
29
accepted Swapping a limit and a $\sup$
Dec
29
accepted LIM is cofinal in ON
Dec
29
asked LIM is cofinal in ON
Dec
29
comment How to evaluate $\int_{1}^{2}\frac{dx}{1+x+\ln x}$?
wolframalpha.com/input/…
Dec
29
comment Showing that this is a group under matrix multiplication
Yes that's enough.
Dec
29
answered Showing that this is a group under matrix multiplication
Dec
29
comment Showing that this is a group under matrix multiplication
For closure: $(AB)^T X AB = B^T A^T X AB = B^T X B = X$.
Dec
27
comment Find $G(n)$ with $n \geq 1$
@qwerty89 Look here: meta.math.stackexchange.com/questions/3286/…
Dec
27
comment Find $G(n)$ with $n \geq 1$
@qwerty89 By accepting he meant that you click on the tick symbol next to the answer you want to accept.
Dec
26
comment Conceptualizing Inclusion Map from Figure Eight to Torus
Dear @Pierre-YvesGaillard: I will see to it.
Dec
26
comment Algebraic Structure of the Rose with Two Petals
@ManuelAraújo He asks for a topological group so it has to have inverses, too. So the previous version of your comment without the "However..." answers the question. : )
Dec
26
comment Algebraic Structure of the Rose with Two Petals
I think in the second paragraph you meant to write "topological group" rather than "topological space".
Dec
26
revised Introductory book on Topology
edited tags