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Closed sets

  1. Feb 15, 2006 #1
    Hi,
    I missed this problem on my test and I need some good pointers about what I did wrong. I want to go talk to my professor, but I feel like a dunce.
    The question is: If A and B are closed in [tex]\mathbb R^1[/tex] show that A X B is closed in [tex]\mathbb R^2[/tex]...or to phrase it more directly, the cross product of A and B is closed in R^2.

    My attempt at this was just SO wrong. I think my professor laughed.
    I know that closed sets contain all of their limit points. I tried to show this by using open balls in A and B..but it just has a Ginormous red X across the thing.
    Please give me the RIGHT start so that when I DO go talk to the prof I'll be a little prepared. I truly do not know why I missed this. I THOUGHT i understood the thing...but I don't.
    CC
     
  2. jcsd
  3. Feb 15, 2006 #2

    AKG

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    Here's one way: let [itex]\pi _i[/itex] be the projection function into the ith component, so [itex]\pi _i(x_1, x_2) = x_1[/itex]. If [itex]U \subset \mathbb{R}[/itex] is open, then [itex]\pi _1 ^{-1}(U) = U \times \mathbb{R}[/itex] and [itex]\pi _2 ^{-1}(U) = \mathbb{R} \times U[/itex]. But both [itex]U \times \mathbb{R}[/itex] and [itex]\mathbb{R} \times U[/itex] are open in [itex]\mathbb{R}^2[/itex], so the projection maps are continuous. Hence, if A and B are closed, then [itex]\pi _1 ^{-1} (A)[/itex] and [itex]\pi _2 ^{-1} (B)[/itex] are closed, so their intersection, A x B, is closed, as required.

    A simpler proof. If A and B are closed, then AC and BC are open in [itex]\mathbb{R}[/itex]. Hence [itex]A^C \times \mathbb{R}[/itex] and [itex]\mathbb{R} \times B^C[/itex] are open in [itex]\mathbb{R}^2[/itex]. Thus, the union of those two sets are open, and thus, the complement of that union is closed. But the complement of that union is precisly A x B, so A x B is closed. To prove that the complement of that union is A x B, it's just basic logic/set theory.
     
  4. Feb 15, 2006 #3
    :blushing: No wonder there's a huge X on this. I made it WAY harder than it is. I think I can get it from here. Thanks
    CC
     
  5. Feb 16, 2006 #4

    HallsofIvy

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    By the way- that is the Cartesian product of sets, not the cross product which applies to vectors.
     
  6. Feb 20, 2006 #5
    Hey,
    I am wondering if this can be done by showing that [tex]\overline{A X B}\subset A X B[/tex].Meaning showing that the closure of A X B is a subset of A X B. We already know that

    [tex]A X B\subset \overline{A X B}[/tex]
    from a theorem in our book.
    Any pointers will be appreciated.
    I think I got it doing it as suggested above.
    Thanks
    CC
     
  7. Feb 20, 2006 #6

    matt grime

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    1. Don't use strict containment when you don't mean to.

    2. Assuming <= is ascii for 'is a subset of' then you've just asked what happens if I show U<=V, where V<=U. Well, obviously U=V in this case, so you can draw your own conclusions.
     
  8. Feb 20, 2006 #7

    quasar987

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    Why is that? I only know of the theorem "A compact, f continuous ==> f(A) compact." but here A and B are not necessarily bounded.


    (Btw - I had a similar question in my test. Used the caracterisation by sequences to prove it. I think this is the most intuitive way.)
     
    Last edited: Feb 20, 2006
  9. Feb 20, 2006 #8
    Let me make sure I stated my question clearly:
    I need to show that[tex]A X B=\overline{A X B}[/tex]
    I know that [tex]A X B\subset \overline{A X B}[/tex]
    How can I show that [tex]\overline{A X B}\subset{A X B}[/tex]
    to prove that if A and B are closed in [tex]\mathbb R[/tex] then A X B is closed in [tex]\mathbb R^2[/tex]?
    This is the method I tried to use on my test. My professor didn't agree with my method and I missed it. I'm wondering if it's even possible.
    CC
     
  10. Feb 20, 2006 #9

    matt grime

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    Definition: a function is continuous iff the inverse image of open sets are open (or closed sets are closed, these are equivalent). It is the only definition of continuity that makes sense, really, in the full generality of topological spaces.
     
  11. Feb 20, 2006 #10

    quasar987

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    Cool. I knew of the "iff the inverse image of open sets are open" definition, but didn't knew it works with closed ones too.
     
  12. Feb 20, 2006 #11
    ok
    so I tried to do it this way:
    I said that: [tex]((A X\mathbb R)\cup(\mathbb R X B))^c[/tex]
    [tex]=(A X \mathbb R)^c \cap (\mathbb R X B)^c[/tex]
    now I need to get that complement inside the parenthesis with the cartesian product and break this down more. From there I went terribly awry. Can anyone help me out?
    CC
     
  13. Feb 20, 2006 #12

    matt grime

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    if you're going to use latex the symbol is \times

    getting the c inside is trivial, and is exactly what it ought to be
     
  14. Feb 20, 2006 #13
    thanks for the code for the big X in there. I'm still learning Latex.
    When I went to see my professor, I discussed with him the method with continuity discussed above. I have also been able to prove it with limit points. He told me that this cannot be proved with the cartesian product complement. I look at it and look at it and I honestly cannot remember how to get the complement of the cartesian product. I remember De Morgan's laws and all of that, but I don't remember ever using set theory for the cartesian product complement. I try to visualize the complement of [tex]A\times B[/tex] and I just can't make sense of it. I want to understand this and see if it works. My professor is adamant that it won't work.
    I know it's trivial to you guys, but I don't know how to do this.
    CC
     
  15. Feb 20, 2006 #14

    AKG

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    It's basic set theory and logic. It's a little disconcerting to hear that your professor says it can't be done using the cartesian product complement.

    [tex](A \times B)^C = \{(x,y)\ |\ x \in A \wedge y \in B\}^C[/tex]

    [tex] = \{(x,y)\ |\ \neg (x \in A \wedge y \in B)\}[/tex]

    [tex] = \{(x,y)\ |\ \neg (x \in A)\ \vee\ \neg (y \in B)\}[/tex]

    [tex] = \{(x,y)\ |\ \neg (x \in A)\}\ \cup \ \{(x,y)\ |\ \neg (y \in B)\}[/tex]

    [tex] = \{(x,y)\ |\ x \notin A\}\ \cup \ \{(x,y)\ |\ y \notin B\}[/tex]

    [tex] = \{(x,y)\ |\ x \in A^C\}\ \cup \ \{(x,y)\ |\ y \in B^C\}[/tex]

    [tex] = (A^C \times \mathbb{R}) \cup \ (\mathbb{R} \times B^C)[/tex]

    A x B is closed iff (A x B)C is open, right? So you know it suffices to prove that (A x B)C is open. And you know that a union of open sets is open, right? So it suffices to prove that each of AC x R and R x BC are open. And you know D is open in X, and E is open in Y, then D x E is open in X x Y, right? So it suffices to show that AC, BC, and R are open in R. But you know that A and B are closed iff their complements are open, and you know that R is open, right? This proof is I would say the easiest, most basic proof, and requires the least knowledge of other theorems. The most "advanced" theorem you need to know for this proof is DeMorgan's Law ;).
     
    Last edited: Feb 20, 2006
  16. Feb 21, 2006 #15

    matt grime

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    My proof for getting the c inside the bracket was somewhat more intuitive than that:

    RxR= (AxR)u(A^cxR)

    so obivoulsy (AxR)^c=A^cxR
     
  17. Feb 21, 2006 #16
    I actually tried to show him the argument above, but he just said no no no...and wrote this down:
    [tex]A^c \times B^c \neq (A\times B)^c[/tex] which wasn't even what I was trying to show him. He likes the proof using the limit point argument. The proof by continuity he's not too fond of. He says he'll have to get back to me on that argument.

    On the above, He said I'd have go through a whole bunch of formulae to convince myself that this argument is not valid. I was CONFUSED because it appeared to me to work. I think he's been teaching this class for a long time and he likes things done the way he likes 'em done.
    He said that if I could convince him that this is true he'd give me a gold star (whoopie...I need extra points to help out my grade....). I guess it's a matter of getting him to listen to what I'm saying instead of assuming I'm wrong before we start. This is a real analysis II class. i don't think he likes looking at logical operators, and the truth be told, it doesn't matter if the proof is right if that's not the way he wants it done.
    I appreciate all of your help on this question. I can see that there are at least 3 ways to do this proof.
    Thanks again.
    CC
     
  18. Feb 21, 2006 #17

    AKG

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    Proposition: If A and B are closed in R, then A x B is closed in R².

    Proof: Suppose A and B are closed in R. To show that A x B is closed in R², it suffices to show that (A x B)C is open in R² (give some justification for this; i.e. it might be how your book defines "closed", or it might be a theorem).

    [itex](A \times B)^C = \{(x,y) \in \mathbb{R}^2\ |\ x \in A \wedge y \in B\}^C[/itex] ... (definition of Cartesian product)

    [itex] = \{(x,y) \in \mathbb{R}^2\ |\ \neg (x \in A \wedge y \in B)\}[/itex] ... (definition of complement)

    [itex] = \{(x,y) \in \mathbb{R}^2\ |\ \neg (x \in A)\ \vee\ \neg (y \in B)\}[/itex] ... (DeMorgan's Law)

    [itex] = \{(x,y) \in \mathbb{R}^2\ |\ \neg (x \in A)\}\ \cup \ \{(x,y) \in \mathbb{R}^2\ |\ \neg (y \in B)\}[/itex] ... (definition of union)

    [itex] = \{(x,y) \in \mathbb{R}^2\ |\ x \in A^C\}\ \cup \ \{(x,y) \in \mathbb{R}^2\ |\ y \in B^C\}[/itex] ... (definition of complement)

    [itex] = (A^C \times \mathbb{R}) \cup \ (\mathbb{R} \times B^C)[/itex] ... (definition of Cartesian product, or, use matt grime's little proof)

    A and B are closed in R, so AC and BC are open in R (again, give justification, which may be "by definition of closed/open"). Since R is open in R, (AC x R) and (R x BC) are open in R x R = R² (by definition of the product topology, or whatever justification you get from your book). The union of two open sets is open (justification), so (AC x R) U (R x BC) is open in R². Since we've proved that this set is equal to (A x B)C, we have shown that (A x B)C is open in R², q.e.d.

    Your job is to fill in the justifications. Pretty much everything in this theorem follows from definition, but your book may define things differently, so things that are definitions in my book may be theorems in yours. From my point of view, the only thing that isn't "by definition" is, as I said, the line that uses DeMorgan's Law. I would recommend showing it to a few different professor first, and have them agree that it is valid. Then, when you show it to your analysis professor, he might get over himself and actually consider the proof you give him.
     
  19. Feb 21, 2006 #18

    AKG

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    Note, the last line of the set theory part of my proof can not use matt grime's proof. If you want to use his proof, remove the last two lines. Then replace them with:

    [itex] = \{(x,y) \in \mathbb{R}^2\ |\ x \in A\}^C\ \cup \ \{(x,y) \in \mathbb{R}^2\ |\ y \in B\}^C[/itex] ... (by definition of complement)

    [itex] = (A \times \mathbb{R})^C\ \cup \ (\mathbb{R} \times B)^C[/itex] ... (definition of cross product)

    [itex] = (A^C \times \mathbb{R})\ \cup \ (\mathbb{R} \times B^C)[/itex] ... (matt grime)

    If you find matt's approach more intuitive, then use it, however, it requires a use of "definition of complement" and "definition of Cartesian product" just as mine does, so whatever intuitive content it has, it also has just as much "nonintuitive" content. Also it requires a little more, so I personally wouldn't use it. In fact, it also requires you to justify that R x R is the disjoint union of AC x R and A x R. It's absolutely obvious, but it would require a fair bit of more work to be absolutely rigorous about it to the degree that your professor can't complain.
     
    Last edited: Feb 21, 2006
  20. Feb 21, 2006 #19

    matt grime

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    I'm puzzled by what you're getting at AKG. All I was talking about was this thing:


    And getting the c inside the bracket is easy because RxR is obviously the union of AxB and A^cxB. Anything of the form (u,v) is either in one or the other. there's nothing tricky in the justiication of that.
     
  21. Feb 21, 2006 #20

    AKG

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    You shouldn't be computing this, since:

    [tex]A \times B \neq (A \times \mathbb{R}) \cup (\mathbb{R} \times B)[/tex].

    Instead:

    [tex]A \times B = (A \times \mathbb{R}) \cap (\mathbb{R} \times B)[/tex]
     
    Last edited: Feb 21, 2006
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