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Some basic questions on the way sets are defined

  1. Jan 25, 2017 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution


    I'm a little confused with how subsets and elements are defined in the case of the given set "A" as it seems to be a set of sets, so I'm going to throw my answers out and would appreciate any guidance on where I'm wrong (if anywhere).

    [itex]A = \Big\{\emptyset, \big\{\emptyset, \{\emptyset\}\big\}\Big\}[/itex]

    "A" has two elements

    a) [itex]\emptyset\subset A[/itex] - True
    b) [itex]\emptyset\in A [/itex] - True
    c) [itex]\{\emptyset\}\subset A [/itex] - Not true
    d) [itex]\{\emptyset\}\in A[/itex] - Not true
    e) [itex]\{\emptyset, \{\emptyset\}\}\subset A[/itex] - True
    f) [itex]\{\emptyset, \{\emptyset\}\}\in A[/itex] - True


    Not sure what's going on here either. I think the issue is in my own flawed understanding of the notation used in sets generally.

    [itex] f : R \rightarrow R[/itex] such that [itex]f(x) = x^{2}[/itex]

    My understanding thus far is that the cartesian product of two sets X and Y is:

    [itex]X \times Y = \{(x,y) : x\in X, y\in Y\}[/itex]

    So in the case of [itex]f(x) = x^2[/itex], we have:

    a) [itex]f((-1,2)) = (-1,2) \times (-1,2) = \big((-1,-1),(-1,2),(2,-1),(2,2)\big)[/itex]

    but then part of me wonders if I've got it all wrong and it should really just be [itex]f((-1,2)) = ((1,4))[/itex] ..??

    And then for part b:

    b) [itex]f((-1,2]) = ... [/itex] ...

    I don't really understand this at all since it has a square bracket which I'm led to believe means it represents a continuous interval of numbers not including that which is on the side of the curled bracket (according to this - interval notation). If that's the case, I don't know how to perform [itex]X \times Y[/itex] in the way I defined above.

    And then we have stuff to do with [itex]f^{-1}[/itex] which is a whole other thing entirely.

    (Just to check - am I right in saying that [itex]f^{-1}[/itex] on a set [itex]Y[/itex] is all the elements [itex]x \in X[/itex] such that [itex]f(x) \in Y[/itex] ??)

    Or in other words: [itex]f^{-1}(Y) = \{ x \in X : f(x) \in Y \}[/itex] - right?

    Hints much appreciated, thanks.
  2. jcsd
  3. Jan 25, 2017 #2


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    You shouldn't post two questions in 1.

    I don't agree with your answers. Which ones are you sure about?
  4. Jan 25, 2017 #3
    To be honest I'm not sure on any of them so far.

    My problem I think lies in that I don't understand the difference between [itex]\{\emptyset\}[/itex] and [itex]\emptyset[/itex]

    To me it seems that one is the empty set, and the other is a set with the empty set in, which is just the empty set, so they're the same. This surely isn't the case because it makes things too trivial, but I don't know how else to interpret those two things.

  5. Jan 25, 2017 #4


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    There is a difference between the element ##a## and the set containing ##a##, denoted by ##\lbrace a \rbrace##. Similarly for the empty set. The set containing the empty set is not empty! It has one member: the empty set.
  6. Jan 25, 2017 #5


    Staff: Mentor

    You could replace the elements of ##A## by some other names, solve the questions and replace them backwards.
  7. Jan 25, 2017 #6
    Excellent, thanks, I think it's all clicked now. Updated answers are below.

    [itex]A = \Big\{\emptyset, \big\{\emptyset, \{\emptyset\}\big\}\Big\}[/itex]

    "A" has two elements

    a) [itex]\emptyset\subset A[/itex] - NOT True because [itex]\emptyset[/itex] is just an element, not a set, and therefore cannot be a subset
    b) [itex]\emptyset\in A [/itex] - True because it is an element of A
    c) [itex]\{\emptyset\}\subset A [/itex] - True since [itex]A = \Big\{\emptyset, \big\{\emptyset, \{\emptyset\}\big\}\Big\} = \{\emptyset\}\cup \Big\{\big\{\emptyset,\{\emptyset\}\big\}\Big\}[/itex], so clearly [itex]\{\emptyset\}\subset A [/itex]
    d) [itex]\{\emptyset\}\in A[/itex] - NOT true, this set is not en element of A.
    e) [itex]\{\emptyset, \{\emptyset\}\}\subset A[/itex] - NOT True, it is only an element.
    f) [itex]\{\emptyset, \{\emptyset\}\}\in A[/itex] - True.
  8. Jan 25, 2017 #7


    Staff: Mentor

    To a): Can you tell all subsets of ##A=\{1,2\}\, ?## How many are there? Why doesn't the argument under c) apply?
  9. Jan 25, 2017 #8
    Hmm, I hope I'm right in saying [itex]\emptyset, \{1\}, \{2\}, \{1,2\}[/itex] are subsets of [itex]\{1,2\}[/itex].

    This now leads me to question my result for part a) however...

    Another way I'm seeing it is that the empty set is {} , hence {}, {1}, {2} and {1, 2} are subsets of {1,2}

    So for { 0, {0,{0}} }, the subsets are {}, {0}, { {0,{0}} } and { 0, {0,{0}} }

    and {} is the empty set ##\emptyset##

    hence ##\emptyset \subset A## ..?
    Last edited: Jan 25, 2017
  10. Jan 25, 2017 #9


    Staff: Mentor

    Yes, the empty set is always a subset, regardless what is in the set. You said it already under c) : ##A=A\cup\{\} \Rightarrow \{\} \subseteq A##.
  11. Jan 25, 2017 #10
    Great stuff, thanks.

    So just to check - the answers I gave for a) to f) were correct apart from part a), which should have been 'true' since ##\emptyset## is always a subset of any set.
  12. Jan 25, 2017 #11


    Staff: Mentor

    If I didn't become confused myself, yes. In the original post c) and e) were wrong. e) could be "repaired" to be true with an additional bracket around: ##\{\{\emptyset, \{\emptyset\}\}\} \subsetneq A\,,## because the element would be then in a set.
  13. Jan 25, 2017 #12
    Great, thanks a lot. All understood now.

    I'll have another look at Q2 tomorrow, probably in a new thread as requested.
  14. Jan 26, 2017 #13
    I think, for d, at least for F(x) is equivalent to

    [tex] \left (-\infty ,0 \right ]\times [0, \infty)[/tex] but dont know how to find f-1, hinki it is an emppty set

    Last edited: Jan 26, 2017
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