Actually, I can't see any real problems here.. about the only issue here is how to translate the defining condition from finite to infinite..

For convenience, let's fix the set of symbols that we are using to be the positive natural numbers N = {1, 2, 3, ... }

(A) An infinite Sudoku has the property that each row, column and box must contain each element of N exactly once

or

(B) An infinite Sudoku has the property that each row, column and box must not have any repeated elements

(For finite Sudoku, these conditions are the same.)

Now, consider the top-left box, which I will call the "base" box. Here is one possibility

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`1 2 4 7 11...`

3 5 8 12..

6 9 13..

10 14..

15..

...

No problem here.. it contains every natural number once each, and has an infinite number of rows and an infinite number of columns - this answers one of your questions, in that the square root of infinity is infinity (at least, the square root of THIS infinity is itself).

Now let's build the other boxes from this one...

Option B is simpler here: just define the (1,2) box (that is, the one that is in row 1, column 2 where here the "rows" and "columns" are the boxes, not the individual rows and columns) by shifting every row up by one (and losing the top row).

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`3 5 8 12..`

6 9 13..

10 14..

15..

...

Then the (1,3) box is obtained by shifting up by one again.

To get the (2,1) box, just shift the COLUMNS left by one, and the (3,1) box similarly and so on.

Then we have an infinite 2d-array of boxes, each box containing natural numbers with no repetitions, such that each individual row/column also contains no repetitions.... but of course each box no longer contains ALL of the natural numbers because we have dropped off the top rows or left-most columns...

For option (A) we have to be a little bit cleverer in order to permute the rows/columns in each box without throwing away any entries. But I think it can be done like this..

For box (1,2), take the rows of the base box in pairs and exchange them

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`row 2`

row 1

row 4

row 3

....

For box (1,3), take the rows of the base box in FOURS and reverse the order of them

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`row 4`

row 3

row 2

row 1

row 8

row 7

...

For box (1,4), take the rows in EIGHTS and in general for box (1,k) take the rows in groups of 2^(k-1) and reverse them.

I *THINK* that this gives a suitable infinite collection of permutations of the rows that does the trick. Then just do the same for columns to get the other boxes.

I don't like to think of how one would actually create a PUZZLE with this as the solution....

Gordon