Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
manjuthottam
Group Title
could someone explain what equinumerous means in mathematical Theoretical course?!
 one year ago
 one year ago
manjuthottam Group Title
could someone explain what equinumerous means in mathematical Theoretical course?!
 one year ago
 one year ago

This Question is Closed

manjuthottam Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
What exactly do you mean by the S\T and T\S? Are those set minuses?
 one year ago

manjuthottam Group TitleBest ResponseYou've already chosen the best response.0
set S without T
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
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\[ g(s)=\begin{cases} f(s)\qquad s\in S\setminus T \\ s\qquad \;\;\;\;\,s\in U \end{cases}\]
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
Did that all make sense?
 one year ago

manjuthottam Group TitleBest ResponseYou've already chosen the best response.0
oh yes thank you!!
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.