**This list is a one-to-one correspondence if everything in the set \(\mathscr{S}\) appears exactly once on the right side.** In other words, if every element of \(\mathscr{S}\) gets a natural number assigned to it (i.e. appears on the list), and it gets only one natural number assigned to it (i.e. it appears only once on the list).

**We will show that any possible listing of the elements of \((0, 1)\) cannot represent a one-to-one correspondence with \(\mathbb{N}\), because it will miss some element of \((0, 1)\).** In fact, we'll imagine we have a listing of the elements of \((0, 1)\), and explicitly construct an element of \((0, 1)\) that is not on that list.

Notice how arbitrary it is. There's certainly no obvious systematic way to proceed, when making such a list. That's okay, though. Maybe someone else came up with a clever algorithm, and generated this list for us.

**But now we'll construct a specific “magic” element of \((0,1)\) that is not on this specific list.** This is how we do it. Our magic element is a decimal between zero and one, so it starts with zero and a decimal point.

**Construction Step 2:** Look at the second number on the list. Look at its hundredths digit (the second one after the decimal point). Whatever that digit is, choose a digit between \(0\) and \(8\) that is different from it, and make that the hundredths digit of the magic number.

In our example, the second number is \(0.0001\). It's hundredths digit is \(0\). So let's choose \(3\). Our magic number now starts "\(0.73…\)".

And we repeat this process, or imagine repeating this process, forever:

At Construction Step \(N\), we look at the \(N\)th number on the list. We look at the digit in the \(N\)th place to the right of the decimal point. Whatever it is, we choose a different digit (between \(0\) and \(8\)), and set that as the digit in the \(N\)th spot to the right of the decimal point of our magic number.

Once we've done this for every \(N\), our magic number is built. Let's call it \(X\). It's a specific real number between zero and one.

**To recap, we started with a (never-ending) list of elements of \((0,1)\). From it, we built a magic decimal \(X\). I claim that this element \(X\) is not on the list (and I'll prove this in a second). Therefore our list cannot be a one-to-one correspondence list -- it didn't actually list all the elements of \((0,1)\).**

Now, sure, we could add \(X\) to the list, maybe put it at the top and bump everything else down. But then we have a new list. With this new list, we could construct a new magic element, call it \(Y\). And by the same logic, \(Y\) is not on the new list. Any list we make will still be missing an element of \((0, 1)\).

**Therefore there can be no one-to-one correspondence between \(\mathbb{N}\) and \((0, 1)\). And this shows that the infinite set is bigger than \(\mathbb{N}\). Note, however, that we could make a one-to-one correspondence between \(\mathbb{N}\) and a subset of \((0,1)\). So \(|(0, 1)| > |\mathbb{N}|\).**

And so if we let the digits of the magic element include nines, in theory we could run into trouble. For example, suppose the list started out like this.

\[0 \leftrightarrow 0.0\\ 1 \leftrightarrow 0.1\]

In theory, if all the rest of the numbers on the list allowed it, we could end up constructing our magic element \(X\) to be \(0.09999999…\). Then in terms of digits, \(X\) differs from everything on the list. However, since \(0.1 = 0.099999…\), our magic element \(X\) is right there on the list, in the second spot!

To avoid this hole in the argument, we just insist that the digits of \(X\) are between zero and eight.