# Size of infinte sets cardinality

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!!

• 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? – bof Mar 26 at 7:40
• same cardinality as N, yes finite sets are countable – Johm Mar 26 at 7:41
• 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 $\mathbb N$.) – bof Mar 26 at 7:44
• 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 – Johm Mar 26 at 7:47
• 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? – bof Mar 26 at 7:49

## 2 Answers

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.
• 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!! – Johm Mar 26 at 7:52
• @Johm I am not talking about the countable subsets of $T$. I am talking about the countable subsets of $R\setminus T$. – 5xum Mar 26 at 7:53
• I was literally misreading the entire thing God. But i understand it – Johm Mar 26 at 8:01

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 $$f\cup g:R\setminus T\to R.$$