More Stupidity with Oleg

Oleg sez:
Umm, the size of an infinite set changes continuously. However if one infinite set contains all of the members of another infinite set AND has members the other does not have...
Dear Oleg,
My methodology is simple. If one infinite set contains all the members of another infinite set, AND it also contains members the other set does not, it has a greater cardinality.
If you know anything about sets, doing that is easy.
And I see Richie is spewing spewing his ignorant "count the letters of the recipe", as if that is all I said was involved...
Oleg sez:
Consider the set {0+x, 1+x, 2+x, 3+x,…}, where x is a real number. When x=0, we have {0,1,2,3,…}. When x=1, we have {1,2,3,4,…}. The great thing about parameter x is that it allows us to continuously interpolate between the two previously considered sets.
The size of a set is “obviously” an integer number and as such it cannot change continuously. Thus, as we increase x continuously from 0 to 1, the size of the set should remain unchanged. So as we arrive at x=1, the set size is still the same as it was when we departed from 0.
So not only do we have a bijective mapping between sets {0,1,2,3,…} and {1,2,3,4,…}, we have a continuous bijective mapping that provides a continuous interpolation between the two sets. Their sizes should certainly be the same.
Umm, the size of an infinite set changes continuously. However if one infinite set contains all of the members of another infinite set AND has members the other does not have...
Dear Oleg,
My methodology is simple. If one infinite set contains all the members of another infinite set, AND it also contains members the other set does not, it has a greater cardinality.
If you know anything about sets, doing that is easy.
And I see Richie is spewing spewing his ignorant "count the letters of the recipe", as if that is all I said was involved...
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