Let's prove that, if a set has size \(n\), then that same set has exactly \(2^n\) subsets.

Take a set of size \(n\). For any of its subsets, we can label items with \(0\)/\(1\) depending on whether or not the item is in the subset or not, and to any such labelling corresponds a single subset. There are \(2^n\) such labellings, hence \(2^n\) subsets.

Do you have an idea for a twitter proof? Let me know in the comments below!Twitter proof:

— Mathspp (@mathsppblog) August 6, 2020

Take a set of size n. For any of its subsets, we can label items with 0/1 depending on whether or not the item is in the subset or not, and to any such labelling corresponds a single subset. There are 2^n such labellings, hence 2^n subsets.https://t.co/5uxKAi7D9T