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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!!

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    $\begingroup$ What does "countable" mean in this question? Some authors use "countable" to mean "has the same cardinality as $\mathbb N$", others use "countable" to mean "has at most the cardinality of $\mathbb N$". Are finite sets "countable" for you? $\endgroup$ – bof Mar 26 at 7:40
  • $\begingroup$ same cardinality as N, yes finite sets are countable $\endgroup$ – Johm Mar 26 at 7:41
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    $\begingroup$ You just gave two contradictory answers to my question: (1) countable sets have the same cardinality as $\mathbb N$ and (2) finite sets are countable. (A finite sets does not have the same cardinality as $\mathbb N$.) $\endgroup$ – bof Mar 26 at 7:44
  • $\begingroup$ yes that is true, what i meant to say was that in this question countable means has the SAME cardinality as N, and finite sets are countable but they do not have same cardinality as N which is true like you said $\endgroup$ – Johm Mar 26 at 7:47
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    $\begingroup$ Your replies continue to contradict themselves. When you say that $T$ is a countable subset of $N$, is it safe to assume that $T$ is infinite? Or could $T$ be a finite set? $\endgroup$ – bof Mar 26 at 7:49
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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.
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  • $\begingroup$ Omg thank you so much!! i get it now i didnt even think about the countable subsets of T since T is countable itself. Appreciate it!! $\endgroup$ – Johm Mar 26 at 7:52
  • $\begingroup$ @Johm I am not talking about the countable subsets of $T$. I am talking about the countable subsets of $R\setminus T$. $\endgroup$ – 5xum Mar 26 at 7:53
  • $\begingroup$ I was literally misreading the entire thing God. But i understand it $\endgroup$ – Johm Mar 26 at 8:01
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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 $$f\cup g:R\setminus T\to R.$$

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