This a strange puzzle about how not all values of infinity are the same.

*If you have a parking lot of infinite width and height, is it possible to have every possible combination of black and white cars, with one combination per column?*

Let us try a problem using hard numbers instead of infinity. It is very possible to fit 4 combinations of 2 cars in a 2x4 matrix:

```
B B W W
```

B W B W

Four possible combinations, in four columns. It is just as possible with a 3x8 matrix:

```
B B B B W W W W
```

B B W W B B W W

B W B W B W B W

As you may have guessed, it is possible to fit any positive real number of cars, n, in 2^{n} columns. Even with 100 rows, you would need 2^{100} columns, which is still finite, despite hardly being able to fit on the Earth*

This begins to fall apart when you have an infinite number of rows. And the question is: is it possible to have every combination, in an infinity*infinity matrix?

You would think: with an infinite number of rows, then surely you would need 2^{infinity} columns; and these would fit snugly within the infinite boundaries of the parking lot, like this:

```
B W W B W B ...
```

W W B W B W ...

B W B B B W ...

B B W B B W ...

W W B W B B ...

W B W B W W ...

| | | | | | \

Suppose that you have managed to organise the cars so every possible combination is there. (The actual grid would be far, far bigger than the 6x6 grid shown there). Well, it's possible, so the answer to the puzzle is yes, then?

However, when you toggle the cars along the diagonal, at (1,1), (2,2), (3,3) and so on, you get a new grid:

```
```**W** W W B W B ...

W **B** B W B W ...

B W **W** B B W ...

B B W **W** B W ...

W W B W **W** B ...

W B W B W **B** ...

| | | | | | \

So in the same number of columns, we have a different set of combinations, accounting for twice as many as we could fit in the grid before.

Why is this the case? As an example, take the first column. It had its first car toggled - and because the grid lists *every* possible combination, there is a column, somewhere, that has the same combination as the first one. But this column has also had one of its cars toggled, so it becomes the same combination as another column; but *that* column has the same combination as another, and so on. Eventually every column will have a new combination, and we end up with a complete grid of different combinations.

Oh dear. I guess the infinitely big grid wasn't big enough after all...

One of the 'solutions' to this puzzle lies in the fact that infinity is not a number, it is a concept, and infinity != 2^{infinity}. You could even go on to say that any number x number grid would be square, and there's no way that you could fit every possible combination of cars in it.

*Even with 1m/car, it would take up 2^{97} km, and the Earth is a meagre 40 000 km in circumference.