Oct 24, 2010 · I remember my professor in college challenging me with this question, which I failed to answer satisfactorily: I know there exists a bijection between the rational numbers and …

The composition of bijections is a bijection. If f f is a bijection, show that h1(x) = 2x h 1 (x) = 2 x is a bijection, and show that h2(x) = x + 2 h 2 (x) = x + 2 is also a bijection. Now we have that g …

Oct 9, 2019 · No, you can't always find a bijection between two uncountable sets. For example, there is never a bijection between any set and its powerset (and sorry, but the standard proof …

If you only have to show that such bijection exists, you can use Cantor-Bernstein theorem and $ (0,1)\subseteq (0,1] \subseteq (0,2)$. See also open and closed intervals have the same …

Now the question remained is how to build a bijection mapping from those three intervels to (0, 1) (0, 1). Or, my method just goes in a wrong direction. Any correct approaches?

May 21, 2025 · 44 "Same cardinality" is defined as meaning there is a bijection. In your vector space example, you were requiring the bijection to be linear. If there is a linear bijection, the …

Having the bijection between (0, 1) (0, 1) and (0, 1)2 (0, 1) 2, we can apply one of the other answers to create a bijection with R2 R 2. The argument easily generalizes to Rn R n.

Apr 15, 2020 · A bijection is different from an isomorphism. Every isomorphism is a bijection (by definition) but the connverse is not neccesarily true. A bijective map f: A → B f: A → B …

Apr 13, 2017 · How does one create an explicit bijection from the reals to the set of all sequences of reals? I know how to make a bijection from $\mathbb R$ to $\mathbb {R \times R}$.

Jan 2, 2025 · I am having some trouble with an intuitive understanding of how we can say two sets equal in cardinality iff there is a bijection between them. In particular, a bijection exists …