What is this analysis exercise talking about?

kindlychung

elimqiu

Let f:S → (0,1) given by  f(x,y) = z with
z=0.A1B1A2B2… whenever x = 0.A1A2… and y=0.B1B2…
Then f is well defined and 1-1, and so |S| ≤ |(0,1)|
But we already know from 1st part of the exercise that |(0,1)|≤ |S|
Thus Schroder-Bernstein Theorem conclude that |S| = |(0,1)| or S ∽ (0,1)

kindlychung

I see.
|S|=|(0,1)|=|R|
What about this hypothesis:
|(a,b)|=|R|=|R^2|=|R^n| ?
If it';s true, how to prove it?

wangyangkee

为老外顶帖，为 elimqiu 顶帖，乐得赚积分，，，