Rudy the Reindeer
Reputation
19,208
Next privilege 20,000 Rep.
Access 'trusted user' tools
 Dec30 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. Dec30 comment LIM is cofinal in ON Where do you use cofinality in the first part of your proof? Dec30 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. Dec30 comment LIM is cofinal in ON I thought the von Neumann universe was the universe of all sets. So it's bigger than ON. Dec30 comment LIM is cofinal in ON Can this proof be modified into not using the von Neumann universe? Dec29 comment LIM is cofinal in ON Thank you, Asaf. Dec29 comment LIM is cofinal in ON Is showing that every cofinal class of ON is proper also a one liner? Dec29 accepted Swapping a limit and a $\sup$ Dec29 accepted LIM is cofinal in ON Dec29 asked LIM is cofinal in ON Dec29 comment How to evaluate $\int_{1}^{2}\frac{dx}{1+x+\ln x}$? wolframalpha.com/input/… Dec29 comment Showing that this is a group under matrix multiplication Yes that's enough. Dec29 answered Showing that this is a group under matrix multiplication Dec29 comment Showing that this is a group under matrix multiplication For closure: $(AB)^T X AB = B^T A^T X AB = B^T X B = X$. Dec27 comment Find $G(n)$ with $n \geq 1$ @qwerty89 Look here: meta.math.stackexchange.com/questions/3286/… Dec27 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. Dec26 comment Conceptualizing Inclusion Map from Figure Eight to Torus Dear @Pierre-YvesGaillard: I will see to it. Dec26 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. : ) Dec26 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". Dec26 revised Introductory book on Topology edited tags