Does this variant of Cantor's diagonal argument work?

Question

I fully realize the following is a less-elegant obfuscation of Cantor's argument, so forgive me. I am still curious if it is otherwise conceptually sound.

1. Make the infinitely-long list alleged to contain every infinitely-long binary sequence, as in the classic argument. Assume each infinite sequence occupies one row.

2. Arrange the list such that it is duplicated in full by optionally taking each column instead of each row—that is, the sequences consisting of the $k^\text{th}$ bits of every row—since if we're asserting there's only one species of infinity, we should be able to put the list in bijection to itself.

   • The most straightforward way I see of doing this is to have the first column and the first row be the same sequence, and likewise for the second column and row, and so forth. The choice of ordering for the list is irrelevant, provided that order is identical on each axis. The resulting table would necessarily be symmetrical about the diagonal.

I think there are lots of ways to kill it from here, but here's one:

3. Notice that one sequence must consist only of infinitely many zeros, and another sequence must consist only of infinitely many ones. Since each sequence appears both horizontally and vertically, it means our zero-sequence must intersect our one-sequence somewhere. Since these two sequences clearly cannot intersect, which requires sharing an element, at least one of them is not on the list, so the list is not complete.

My question is whether this is substantively the same as Cantor's argument, more or less, or if I've introduced any additional suspect assumptions without realizing it which could break the argument.

Answer 1

This argument just doesn't work at all. You have not justified your claim in step 2 that you can arrange both that every sequence appears in a row and that every sequence appears in a column. There is no reason to expect this to be true.

It is instructive to consider what would happen if you were instead only trying to list all the sequences that have only finitely many $1$s. In this case, the set of such sequences is countable, so it is possible to do so! However, it is not possible to do so such that the columns also include every such sequence. For if you could, then one of the columns (say column $n$) would be all $0$s, but then that would mean every row has a $0$ as its $n$th entry, so your list includes no sequences whose $n$th term is $1$.