## manjuthottam Group Title could someone explain what equinumerous means in mathematical Theoretical course?! one year ago one year ago

1. manjuthottam Group Title

the definition states " two sets S and T are called equinumerous, and we write S ~ T, if there exists a bijective function from S onto T". Bijective definition is "a function is bijective if it is surjective and injective". I'm asked to "Prove that if (S \ T) is equinumerous to (T \ S), then S is equinumerous to T" how do i do that?

2. KingGeorge Group Title

What exactly do you mean by the S\T and T\S? Are those set minuses?

3. manjuthottam Group Title

yes

4. manjuthottam Group Title

set S without T

5. KingGeorge Group Title

So there are the same number of elements in S that aren't in T as there are elements in T that aren't in S. Let $$U=S\cap T$$. So $$S\setminus T=S\setminus (S\cap T)$$. Then $$S=S\setminus T +U$$. Now let $$f:S\setminus T\to T\setminus S$$ such that $$f$$ is a bijection. Now define$g:S\setminus T+U\longrightarrow T\setminus S+U$by$g(s)=\begin{cases} f(s)\qquad s\in S\setminus T \\ s\qquad \;\;\;\;\,s\in U \end{cases}$

6. KingGeorge Group Title

Note that if we restrict $$g$$ to only the domain $$U$$, we get a bijective function since $$g(g(s))=s$$. I.e., $$g$$ has an inverse. Similarly, if we restrict $$g$$ to only the domain $$S\setminus T$$, then we have a bijective function since $$f$$ is bijective. So $$g$$ is bijective over its whole domain, and is therefore a bijective function. Finally, since $$S\setminus T+U=S$$ and $$T\setminus S+U=T$$, $$S$$ and $$T$$ are equinumerous.

7. KingGeorge Group Title

Did that all make sense?

8. manjuthottam Group Title

oh yes thank you!!