Half of the multiples of 3 are also multiples of 2. Those can map to themselves. The multiples of 3 that are not also multiples of 2 can map to even numbers between the adjacent two multiples of 3. For instance, 6 maps to itself and 12 maps to itself. 9 can map to a number between 6 and 12; let’s pick 8. That leaves 10 unaccounted for with no multiples of 3 going back to deal with it later; therefore, there are more multiples of 2 than of 3.
Essentially, you’ve walked up the natural numbers in order and noted that you encounter more multiples of 2 than multiples of 3. But there’s no reason to privilege that particular way of encountering elements of the two sets.
For instance, instead of mapping multiples of 3 to a close multiple of 2, we could map each multiple of 3 to two-thirds of itself. Then every multiple of 2 is accounted for, and there are exactly as many multiples of 2 as of 3. Or we could map even multiples of 3 to one third their value, and then the the odd multiples of 3 are unaccounted for, and we have more multiples of 3 than of 2.
Your intuition seems to correspond to the following: if, in any large enough but finite segment of the number line, there are more members of set A than of set B; then |A|>|B|.
The main problem with this is that it contradicts another intuition (which I hope you share, please say so explicitly if you don’t): if you take a set A, and map it one-to-one to a different set B (a complete and reversible mapping) - then the two sets are equal in size. After all, in some sense we’re just renaming the set members. Anything we can say about set B, we can also say about set A by replacing references to members of B with members of A using our mapping.
But I can build such a mapping between multiples of 2 and of 3: for every integer x, map 2x to 3x. This implies the two sets are equal contradicting your intuition.
Half of the multiples of 3 are also multiples of 2. Those can map to themselves. The multiples of 3 that are not also multiples of 2 can map to even numbers between the adjacent two multiples of 3. For instance, 6 maps to itself and 12 maps to itself. 9 can map to a number between 6 and 12; let’s pick 8. That leaves 10 unaccounted for with no multiples of 3 going back to deal with it later; therefore, there are more multiples of 2 than of 3.
Essentially, you’ve walked up the natural numbers in order and noted that you encounter more multiples of 2 than multiples of 3. But there’s no reason to privilege that particular way of encountering elements of the two sets.
For instance, instead of mapping multiples of 3 to a close multiple of 2, we could map each multiple of 3 to two-thirds of itself. Then every multiple of 2 is accounted for, and there are exactly as many multiples of 2 as of 3. Or we could map even multiples of 3 to one third their value, and then the the odd multiples of 3 are unaccounted for, and we have more multiples of 3 than of 2.
Your intuition seems to correspond to the following: if, in any large enough but finite segment of the number line, there are more members of set A than of set B; then |A|>|B|.
The main problem with this is that it contradicts another intuition (which I hope you share, please say so explicitly if you don’t): if you take a set A, and map it one-to-one to a different set B (a complete and reversible mapping) - then the two sets are equal in size. After all, in some sense we’re just renaming the set members. Anything we can say about set B, we can also say about set A by replacing references to members of B with members of A using our mapping.
But I can build such a mapping between multiples of 2 and of 3: for every integer x, map 2x to 3x. This implies the two sets are equal contradicting your intuition.