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Math Help - Topology beginer some help with proofs please....

  1. #16
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    Re: Topology beginer some help with proofs please....

    Quote Originally Posted by Bwts View Post
    If B \subset \bigcap_{\alpha \in I} then for every x \in B, x \in \bigcap_{\alpha \in I} A_\alpha

    If x \in \bigcap_{\alpha \in I} A_\alpha then for every \beta \in I, x \in A_\beta
    That's correct. Let's call these statements (1) and (2).

    Quote Originally Posted by Bwts View Post
    So x \in B if and only if B \subset A_\beta
    This is obviously false and in no way follows from the previous statements. If either the right- or the left-hand side of the original claim is true, then B\subset A_\beta for all \beta. However, x can be arbitrary; it can belong to B or not.

    Combine (1) and (2) and write B \subset \bigcap_{\alpha \in I} A_\alpha without using \subset or \bigcap. Then do the same with the right-hand side, i.e., "for each \beta \in I, B \subset A_\alpha." Compare the results.
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  2. #17
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    Re: Topology beginer some help with proofs please....

    Prove B \subset \bigcap_{\alpha \in I} A_\alpha if and only if for each \beta \in I, B \subset A_\alpha.

    If B \subset \bigcap_{\alpha \in I} A_\alpha then for every x \in B, x \in \bigcap_{\alpha \in I} A_\alpha (1)

    If x \in \bigcap_{\alpha \in I} A_\alpha then for every \beta \in I, x \in A_\beta (2)

    So for every x \in B for each \beta \in I, x \in A_\beta (3)

    R. H. S.

    For each \beta \in I for each x \in B, x \in A_\alpha (4)

    Are (3) & (4) equivalent (they look like theyare to me but the indices are throwing me)?
    Last edited by Bwts; July 26th 2012 at 08:09 PM.
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  3. #18
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    Re: Topology beginer some help with proofs please....

    i, too, am confused by the indices. i think it should read:

    B \subset \bigcap_{\alpha \in I} A_\alpha if and only for each  \beta \in I, B \subset A_\beta

    or:

    B \subset \bigcap_{\alpha \in I} A_\alpha if and only for each  \alpha \in I, B \subset A_\alpha

    which i find preferable.

    in plain english, what is being said is:

    "B is contained in the mutual intersection of all the A's, if and only if B is contained in each A".

    now one direction of this equivalence is obvious:

    if B is in the intersection of all of the A's, B is obviously contained by each and every single A, since the intersection of a family of sets is a subset of each set (this follows directly from the definition of intersection).

    the other direction is a little more subtle. basically, you are generalizing to an infinite family the statement:

    [(B ⊂ A1) & (B ⊂ A2)] → (B ⊂ (A1∩A2))

    the "&" is most important, here.
    Thanks from Bwts
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  4. #19
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    Re: Topology beginer some help with proofs please....

    Quote Originally Posted by Bwts View Post
    Prove B \subset \bigcap_{\alpha \in I} A_\alpha if and only if for each \beta \in I, B \subset A_\alpha.
    Should be  A_\beta .

    Quote Originally Posted by Bwts View Post
    If B \subset \bigcap_{\alpha \in I} A_\alpha then for every x \in B, x \in \bigcap_{\alpha \in I} A_\alpha (1)

    If x \in \bigcap_{\alpha \in I} A_\alpha then for every \beta \in I, x \in A_\beta (2)

    So for every x \in B for each \beta \in I, x \in A_\beta (3)
    Yes, but I would replace "If... then..." in (1) and (2) by "iff."

    Quote Originally Posted by Bwts View Post
    R. H. S.

    For each \beta \in I for each x \in B, x \in A_\alpha (4)
    Should be A_\beta .

    Quote Originally Posted by Bwts View Post
    Are (3) & (4) equivalent (they look like theyare to me but the indices are throwing me)?
    Yes because two consecutive "for all"s can be swapped.
    Thanks from Bwts
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  5. #20
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    Re: Topology beginer some help with proofs please....

    Thanks yes I transcribed the original question wrong and have propagated the error for the last 2 days (I would say that'll teach me but it probably wont )

    OK so lets see if I'm getting the hang of this,

    Let \{ A \}_{\alpha \on I} be an indexed family of subsets of a set S and let B \subset S

    Prove that \bigcup_{\alpha \in I} A_\alpha \subset B iff for each \beta \in I, A_\beta \subset B

    \bigcup_{\alpha \in I} A_\alpha \subset B iff for every x \in \bigcup_{\alpha \in I} A_\alpha, x \in B (1)

    x \in \bigcup_{\alpha \in I} A_\alpha iff for a least one \beta \in I, x \in A_\beta (2)

    So \bigcup_{\alpha \in I} A_\alpha \subset B iff for each \beta \in I, A_\beta \subset B (as we dont know which A_\beta x is in)?

    Doesnt look right to me, is it?
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  6. #21
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    Re: Topology beginer some help with proofs please....

    Quote Originally Posted by Bwts View Post
    Let \{ A \}_{\alpha \on I} be an indexed family of subsets of a set S and let B \subset S

    Prove that \bigcup_{\alpha \in I} A_\alpha \subset B iff for each \beta \in I, A_\beta \subset B

    \bigcup_{\alpha \in I} A_\alpha \subset B iff for every x \in \bigcup_{\alpha \in I} A_\alpha, x \in B (1)

    x \in \bigcup_{\alpha \in I} A_\alpha iff for a least one \beta \in I, x \in A_\beta (2)
    (1) and (2) are correct.

    Quote Originally Posted by Bwts View Post
    So \bigcup_{\alpha \in I} A_\alpha \subset B iff for each \beta \in I, A_\beta \subset B (as we dont know which A_\beta x is in)?

    Doesnt look right to me, is it?
    This is what you need to prove, so it is right. However, you don't get it immediately. Statements (1) and (2) are combined into

    For every x, if there exists a \beta\in I such that x\in A_\beta, then x\in B.

    If we write \forall for "for all" and \exists for "there exists," then this is

    \forall x.\,((\exists\beta\in I.\,x\in A_\beta)\to x\in B). (3)

    What you need to prove is (after expanding \subset)

    \forall\beta\forall x.\,(x\in A_\beta\to x\in B). (4)

    It is important that in (3), the whole \exists\beta\in I.\,x\in A_\beta is in the premise of the implication. Now, one of logical laws says that (\exists x\,P(x))\to Q is equivalent to \forall x.\,(P(x)\to Q), where P(x) is any property of x and Q is any proposition that does not depend on x. (Make sure it is obvious to you,) Apply it to (3) and then swap the consecutive \forall to get (4).
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  7. #22
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    Re: Topology beginer some help with proofs please....

    i don't think that reducing the statement to the equivalence of logical forms involving universal and existential quantifiers is particularly enlightening, here (although, admittedly, that is *one* way to proceed).

    suppose that \bigcup_{\alpha \in I} A_\alpha \subset B, and choose an arbitrary \beta \in I.

    now pick any x \in A_\beta. since x \in A_\beta,\ x \in \bigcup_{\alpha \in I} A_\alpha.

    since \bigcup_{\alpha \in I} A_\alpha \subset B,\ x\in B. this shows that A_\beta \subset B for any (thus all) \beta \in I.

    in the other direction, we suppose that for all \beta \in I, A_\beta \subset B.

    now we pick an (arbitrary) x \in \bigcup_{\alpha \in I} A_\alpha.

    by the *definition* of union, x \in A_\beta for some \beta \in I.

    thus x \in B, since for any \beta \in I,\ A_\beta \subset B, which shows that \bigcup_{\alpha \in I} A_\alpha \subset B.
    Last edited by Deveno; July 27th 2012 at 12:04 PM.
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  8. #23
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    Re: Topology beginer some help with proofs please....

    OK thanks to you both. I'll try a few more excercises over the weekend & will probably be back with more questions & questionable proofs.

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  9. #24
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    Re: Topology beginer some help with proofs please....

    OK I'm really struggling here...

    1. \{A\}_{\alpha \in I} and \{B\}_{\alpha \in I} are 2 indexed families of subsets in S.

    a) Prove for each \beta \in I, A_\beta \subset \bigcup_{\alpha \in I} A_\alpha

    My attempt...

    Let x \in A_\beta for any \beta \in I

    If x \in \bigcup_{\alpha \in I} A_\alpha then for some (at least one) \alpha \in I, x \in A_\alpha

    ............lost it at this point.......

    2. For each \beta \in I, A_\beta \subset B_\beta

    Prove \bigcup_{\alpha \in I} A_\alpha \subset \bigcup_{\alpha \in I} B_\alpha

    Attempt one...

    If \bigcup_{\alpha \in I} A_\alpha \subset \bigcup_{\alpha \in I} B_\alpha then for all x \in \bigcup_{\alpha \in I} A_\alpha, x \in \bigcup_{\alpha \in I} B_\alpha (1)

    If x \in \bigcup_{\alpha \in I} A_\alpha then for some \alpha \in I, x \in A_\alpha and x \in B_\alpha (2)

    ............lost it at this point.......

    Attempt two...

    Let \beta \in I and x \in A_\beta then x \in \bigcup_{\beta \in I} A_\beta

    As A_\beta \subset B_\beta then x \in B_\beta so x \in \bigcup_{\beta \in I} B_\beta


    ....err yeah lost it at this point......

    I'm not having big issues with the other proofs I'm trying (besides have to think about them for a long time) but for some reason I cant get my head around these changes of indicies.
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  10. #25
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    Re: Topology beginer some help with proofs please....

    Quote Originally Posted by Bwts View Post

    2. For each \beta \in I, A_\beta \subset B_\beta

    Prove \bigcup_{\alpha \in I} A_\alpha \subset \bigcup_{\alpha \in I} B_\alpha
    ..............

    Attempt two...

    Let \beta \in I and x \in A_\beta then x \in \bigcup_{\beta \in I} A_\beta

    As A_\beta \subset B_\beta then x \in B_\beta so x \in \bigcup_{\beta \in I} B_\beta
    So does this directly imply that \bigcup_{\beta \in I} A_\beta \subset \bigcup_{\beta \in I} B_\beta?

    If so how do I go from \beta to \alpha?

    Just because \beta \in I and \alpha \in I doesnt make \beta = \alpha does it? What am I missing here?
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  11. #26
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    Re: Topology beginer some help with proofs please....

    It is not recommended to ask more than two questions in one thread. Otherwise, threads become difficult to navigate. You should create a new thread and copy your attempts there.

    Quote Originally Posted by Bwts View Post
    a) Prove for each \beta \in I, A_\beta \subset \bigcup_{\alpha \in I} A_\alpha

    My attempt...

    Let x \in A_\beta for any \beta \in I

    If x \in \bigcup_{\alpha \in I} A_\alpha then for some (at least one) \alpha \in I, x \in A_\alpha
    What you must mean is, "To prove x \in \bigcup_{\alpha \in I} A_\alpha it is sufficient to show that for some (at least one) \alpha \in I, x \in A_\alpha." So, is it really difficult to show this when you assumed that x \in A_\beta for some \beta \in I?

    It's useless to say "If x \in \bigcup_{\alpha \in I} A_\alpha." Suppose the claim that starts with this "If" is true. How are you going to use it? To use "If A, then B," one needs to provide a proof of A; then he/she will obtain a proof of B. But you can't provide a proof of the premise because the premise in your claim is exactly what needs to be ultimately proven!

    This reminds me the following joke.
    As an experiment, an engineer, a physicist and a mathematician are placed in separate rooms and left with a can of food but no can-opener. A day later, the rooms are opened one by one.

    In the first room, the engineer is snoring, with a battered, opened and emptied can. When asked, he explains that when he got hungry, he beat the can to its failure point.

    In the second room, the physicist is seen mouthing equations, with a can popped open beside him. When asked, he explains that when he got hungry, he examined the stress points of the can, applied pressure, and 'pop'!

    In the third room, the mathematician is found sweating, and mumbling to himself, 'Assume the can is open, assume the can is open...'
    Here, the goal is to get the can opened. What's the use to assume it? One needs to achieve/produce it!

    For another example, a proven implication "If A, then B" is similar to a store that sells B's for the price A. If you have an A in your pocket, you can come to the store, pay A and get a B. This is different from the situation where you need to prove "If A, then B." This situation is similar to a workshop that produces B's from A's. If somebody gives you an A, you need to know how to manufacture a B from it.

    I recommend that you review how to prove claims of the form "For all x, P(x)," "There exists an x such that P(x)," "If A, then B," "A and B," "A or B" and "not A." Also review how to use such claims when they are already proven.
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