Determine a subset for the function of X -> P(X)

  • #1

Homework Statement



Suppose X is a non-empty set and f:X->power set of X (ie P(X)) is defined f(x) = X/x. Consider the subset Yf = {x in X such that x not in f(x)} of X. Determine Yf for the particular f we have just described.

Homework Equations





The Attempt at a Solution



Any suggestions would be helpful because I really don't know where to start on this problem...
 

Answers and Replies

  • #2
Dick
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This really isn't a profound question. It looks like it's just testing if you can read the notation. Pick X={1,2,3}. What's f(1)?
 
  • #3
f(1) in that case would be {2,3} right? but how does that translate to Yf?
 
  • #4
Dick
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f(1) in that case would be {2,3} right? but how does that translate to Yf?

Right. Is 1 in Y_f?
 
  • #6
Dick
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no, so is Yf just X/x ?

Read the definition of Y_f again. Aloud. I don't want to know what Y_f is yet, I just want to know if 1 is in it. No is the wrong answer. Why?
 
  • #7
no 1 isn't in it because it's x in X such that x not in f(x) and 1 is not in f(1)
 
  • #8
Dick
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no 1 isn't in it because it's x in X such that x not in f(x) and 1 is not in f(1)

If 1 is not in f(1), doesn't that mean 1 IS in Y_f?
 
  • #9
I get that for f(1), 2 is in Y_f and 3 is in Y_f but 1 is not.
for f(2), 1 is in Y_f and 3 is in Y_f but 2 is not.
so for f(n) on a set of size t Y_f will have all numbers in the set 1 through t not including n in it. is that how you'd say that or is there a more explicit formula?
 
  • #10
Dick
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I get that for f(1), 2 is in Y_f and 3 is in Y_f but 1 is not.
for f(2), 1 is in Y_f and 3 is in Y_f but 2 is not.
so for f(n) on a set of size t Y_f will have all numbers in the set 1 through t not including n in it. is that how you'd say that or is there a more explicit formula?

Ach. I think you are getting this wrong. 1 ISN'T in f(1). Doesn't that tell you 1 IS in Y_f? Or did you state the question wrong?
 
Last edited:
  • #11
I don't know I thought f just took n to a set with all elements of X not including n because Y_f is x not in f(x)
 
  • #12
Dick
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I don't know I thought f just took n to a set with all elements of X not including n because Y_f is x not in f(x)

That's really confusing. Let's go back. If X={1,2,3} is 1 in Y_f? I don't think we really ever got that one right.
 
  • #14
Dick
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no it's not

Yes, it is. Yf = {x in X such that x not in f(x)}. 1 is in X but it's NOT in f(1)={2,3}. So 1 IS in Y_f.
 
  • #15
ohhhh. ok so Y_f is whatever value you plug in for x?
 
  • #16
Dick
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ohhhh. ok so Y_f is whatever value you plug in for x?

Hmm. What Y_f is actually doesn't depend on what x is. x is what's called a dummy variable. Work it out for X={1,2,3} and tell me what members of X are in Y_f.
 
  • #17
{1,2,3} are in Y_f because f(1) implies 1 is in it, f(2) implies 2 is in it and f(3) implies 3 is in it right?
 
  • #18
Dick
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{1,2,3} are in Y_f because f(1) implies 1 is in it, f(2) implies 2 is in it and f(3) implies 3 is in it right?

Right. So Y_f=X={1,2,3}. Does the same thing work for every set?
 
  • #20
Dick
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  • #22
Dick
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so Y_f is just X?

What do YOU think? That's a lot more important than what I think.
 
  • #24
Dick
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  • #25
ok. Thank you very much!
 

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