Size of infinte sets cardinality

Question

The question is as follows:

Prove that if R is uncountable and T is a countable subset of R, then the cardinality of R\T is the same as the cardinality of R.

What i have:

I know that R is uncountable so it has a countable subset (this is a theorem of uncountable sets). Let T be this subset so T has the same cardinality as the set of natural numbers(by the definition of T being countable). My intution is telling me that we will have to use the cantor bernstein theorem to prove they have same cardinalities. So for that the first thing i did showed was that |R\T| <= |R| (pretty clear as its R\T, |R| means cardinality of R).i got a bit confused while trying to show that |R| <= |R\T|. Maybe we can show this by defining a bijection f from R --> R\T, such that f(x) = x when x is in R\T, but i dont know what to do if x is in R and not in R\T, if i can define that function then i can conclude |R| <= |R\T|, then use the cantor-bernstein theorem and then im done. Or maybe im doing this all wrong i cant think of any other way Any help is much appreciated!!

Answer 1

I think you can find a quite nice bijection between $R$ and $R\setminus T$, actually. I would go at it like so:

1. Prove that $R\setminus T$ is uncountable
2. Conclude that there must exist some countable $T_1\subseteq R\setminus T$, with $|T_1|=|T|$.
3. From there, we can find some countable $T_2\subseteq R\setminus (T\cup T_1)$, and continue from there to find countable $T_1,T_2,T_3,\dots T_n,\dots$, such that for each $n$, we have $T_n\subseteq R\setminus(T\cup T_1\cup\cdots\cup T_{n-1})$
4. Now, we can map the values of $x\in T$ to elements in $T_1$, elements in $T_1$ to $T_2$, and so on. All values not in any of $T_i$ get mapped to themselves.

Answer 2

Let $T$ be a countable subset of an uncountable set $R$.

Since $R$ is uncountable and $T$ is countable, it follows that $R\setminus T$ is uncountable. Therefore, there is a countably infinite set $S\subseteq R\setminus T$. Then $S\cup T$ is also a countably infinite set. Therefore there is a bijection $$f:S\to S\cup T.$$ Of course there is a trivial bijection $$g:R\setminus(S\cup T)\to R\setminus(S\cup T).$$ The union of these two bijections is a bijection $$f\cup g:R\setminus T\to R.$$