Dec 27, 2014 · 3 In Royden's book Real Analysis, page 13, he writes that "We call two sets A and b equipotent provided there is a one-to-one mapping f f from A onto B Equipotence defines a …

We know by definition that if a bijection between two sets A and B exists,then A and B are equivalent.The book I was reading took the function f:Z→N f(x)= -2x,for x<0 2x+1,for x=>0 w...

Answer : Yes Hint : Since $\mathbb {R}$ is equipotent to $]0,1 [$, you just have to prove $]0,1 [$ is equipotent to $]0,1 [^2$. Now you can use decimal expansion and the same kind of trickery you …

I know that $\mathbb N^2$ is equipotent to $\mathbb N$ (By drawing zig-zag path to join all the points on xy-plane). Is this method available to prove $\mathbb R^2 $ equipotent to $\mathbb R$?

Feb 7, 2022 · An infinite set is actually often defined as a set that is equipotent to some proper subset. When dealing with foundational questions like this, you need to be very precise with the definitions …

Nov 28, 2014 · Then there is a set $B$ such that $A$ is equipotent to $B\times\mathbb N$. Assuming this lemma, we only need to prove your property when $A$ is in fact $B\times \mathbb N$.

Jun 7, 2015 · Prove that sets being equipotent is an equivalence relation Ask Question Asked 10 years, 4 months ago Modified 5 years, 6 months ago

Nov 30, 2016 · @DonAntonio: Yes, I can use it. But how do I use CSB without giving two injections?

Feb 2, 2021 · This is a basic question regarding the definition of cardinal numbers using the equipotent relation or equinumericity . (1) From what I have understood so far, the strategy is to define the …

But the set of all finite subsets of $E$ is equipotent to $E$. So its complement in $\mathfrak {P} (E)$ has to be equipotent to $\mathfrak {P} (E)$ by Cantor's theorem.