Cantor proved that there are no bijective functions from the natural numbers to the real numbers, and thus they have different cardinalities. His diagonal proof shows there’s an injective function from \(\mathbb{N} \to \mathbb{R}\) but no bijective function can exist, and thus
\(|\mathbb{N}| < |\mathbb{R}|\). Injection but no bijection implying a strict inequality is a definition of cardinality. \(\mathbb{R}\) is thus a “larger” infinity.
The broad idea is to “list” all of the naturals (where listing means a finite description clearly showing how one would enumerate them in an order if one had infinite time), show each one maps to a different real (showing an injection), and then showing a new real can be produced that’s not on the list already. Since no natural maps to it (since they’re all already “listed”), there can’t be a surjection and likewise no bijection.
Let: \(f: \mathbb{N} \to [0,1] \) be any injective function. Letting \(f\) be any injective function will show no surjection (and thus no bijection) exists. Since it’s any injective function, pick arbitrary reals, and put them into infinite decimal expansion form:
\[\begin{array}{c} \begin{array}{c ccccccccccc} n & & & & & & f(n) & & & & & & & & & \end{array} \\ \begin{array}{c | ccccccccccc} \hline 1 & 0 & . & 3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & ... \\ 2 & 0 & . & 4 & 6 & 6 & 9 & 2 & 0 & 1 & 6 & ... \\ 3 & 0 & . & 2 & 5 & 0 & 2 & 9 & 0 & 7 & 8 & ... \\ 4 & 0 & . & 2 & 7 & 1 & 8 & 2 & 8 & 1 & 8 & ... \\ 5 & 0 & . & 5 & 9 & 9 & 0 & 4 & 1 & 6 & 7 & ... \\ ⋮ & ⋮ \end{array} \end{array}\]While the first column is all of the naturals, the second column is not all of the reals. Why? Because any attempt to list them shows a real will be missed (that is not true for listing the naturals; we know every one will be listed, clearly). To create a real not on “the list”, simply add \(1\) to the first digit past the decimal for row one, the second digit for row two, and so on (letting \(9+1)\) wrap around to \(0\):
\[\begin{array}{c} \begin{array}{c ccccccccccc} n & & & & & & f(n) & & & & & & & & & \end{array} \\ \begin{array}{c | ccccccccccc} \hline 1 & 0 & . & \textbf{3} & 1 & 4 & 1 & 5 & 9 & 2 & 6 & ... \\ 2 & 0 & . & 4 & \textbf{6} & 6 & 9 & 2 & 0 & 1 & 6 & ... \\ 3 & 0 & . & 2 & 5 & \textbf{0} & 2 & 9 & 0 & 7 & 8 & ... \\ 4 & 0 & . & 2 & 7 & 1 & \textbf{8} & 2 & 8 & 1 & 8 & ... \\ 5 & 0 & . & 5 & 9 & 9 & 0 & \textbf{4} & 1 & 6 & 7 & ... \\ ⋮ & ⋮ \end{array} \end{array}\]Doing so, for any codomain for this injective function, with any attempt to list it, will always produce a new real not previously on the list. That’s because it will differ from each number already “on the list” by one digit. Therefore, while a injection exists, no surjection will ever exist. Keep in mind, no surjection exists when letting \(f\) be an injective function from the naturals. Letting \(f\) be such a function is precisely the assumption we want to make. We have to have an injection to even attempt a bijection. And that we have one or many, and then no surjections, proves by definition \(|\mathbb{N}| < |\mathbb{R}|\). Hence the reals are a greater infinity.