The other is to construct its inverse explicitly, thereby showing that it has an inverse and hence that it must be a bijection. You could take that approach to this problem as well:

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 …

Apr 15, 2020 · 2 A bijection is an isomorphism in the category of Sets. When the word "isomorphism" is used, it is always referred to the category you are working in. I will list some …

A continuous bijection can't exists because $ [0,1]$ is a compact set and continuous functions send compacts in compacts. You can look for a non-continuous bijection, that exists because …

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 …

Having the bijection between $ (0,1)$ and $ (0,1)^2$, we can apply one of the other answers to create a bijection with $\mathbb {R}^2$. The argument easily generalizes to $\mathbb {R}^n$.

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 …

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 $\mathbb {R \times R}$.

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 …

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