Verified answer

Ben: Showing that S is uncountable doesn't get us very far. If you assume the continuum hypothesis it shows that |S| >= |R|. Otherwise it's not clear what we gain.

Coco: This is very similar to the proof that R is equinumerous with the set S' of functions f: T -> {0.1}. You'll recall we proved this is follows. Enumerate the rational numbers as q_0, q_1, ... Let r be a real number. The function f_r: N -> {0,1} which returns 0 if q_n < r and 1 if q_n >=r is distinct for each real number r. Hence |R| <= |S'|. Conversely, for each distinct function f in S', the real number f(0)/(3^0) + f(1)/(2^1) + ... + f(n)/(3^n) + ... is distinct. (To prove this, just note that the denominator is sufficiently large that if two functions disagree on some first n, the second will never "catch up". You can make this rigorous using geometric series.) Hence |R| >= |S'|, so |R| = |S'|.

Almost the same proof works for S. Since S contains S', our injection of R into S' also injects R into S, and hence |R| <= |S|. Our injection from S' into R is no longer an injection, because the constant in the denominator is not large enough. Make it bigger (100 should work; I'm too lazy to calculate how big it has to be) and the proof goes through as before.

Hope this helps!

Well I assume what you mean to say is to show that the set |S| is uncountable.

Suppose that you did have a way of counting all the functions f:N -> T. That is, we have some set F = {f_1, f_2, f_3, ..., f_i, ...}

where for each f, if n is in N, we have

f_i(n) = 1, 2, or 3.

which contains each possible function.

We can then construct the following function g:N -> T.

For any i in N we say that:

if f_i(i) = 1, then g(i) = 2

if f_i(i) = 2, then g(i) = 3

if f_i(i) = 3, then g(i) = 1.

It follows that for any i, g(i)≠f_i(i). Thus, g(n) is not contained in F, which means that F cannot be constructed properly. Thus, S is uncountable.

right here f(a,b) = 2^a x 3^b? and enable A = { 0,a million,2,3,4} x { 0,a million,2) and enable B = { n: n is a good element of a hundred and forty four}. as a result a million) The functionality is onto and 2) the functionality is one-to-one.