What you need to show is that for every number of the form 4m + 1, that this number is also of the form 2n - 1, such that m, n are integers. This is simple enough, just set:

2n - 1 = 4m + 1

Given an m we should be able to find an n, and we can:

n = 2m + 1

(Since m is an integer, so is 2m. Since 2m is an integer so is 2m + 1. Thus n is also an integer.)

So every element of the set is an element of .

In the event you need this to be a proper inclusion, note that not every element of is in because:

4m + 1 = 2n - 1

gives

m = (1/2)(n - 1)

Which means there are values of n such that m is not an integer. So we have:

but not the other way around. Hence this is a proper inclusion.

-Dan