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:

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 is …

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 natural …

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 cardinality …

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 dimension is …

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 $\mathbb {R}^n$.

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?

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 categories including their …

Dec 12, 2014 · For example, if the sets are groups, then an isomorphism is a bijection that preserves the operation in the groups: $\varphi (ab) = \varphi (a)\varphi (b)$. As another example, if the sets are …

Jan 21, 2025 · To my understanding, an isomorphism is a bijection that also preserves a specific structure, such as algebraic or geometric operations. While every isomorphism is a bijection, not all …