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Math Help - Union of indexed sets

  1. #1
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    Union of indexed sets

    "\text{Given } \{A_\alpha\}_{\alpha \in I}, J\subset I, \text{show } \bigcup_{\alpha \in J} A_\alpha \subset\bigcup_{\alpha \in I} A_\alpha."

    Here is what I did:
    1. \bigcup_{\alpha \in J} A_\alpha = \{x \in S: x \in A_\beta \text{ for some } \beta \in J\}, (*)
    \bigcup_{\alpha \in I} A_\alpha = \{x \in S: x \in A_\beta \text{ for some } \beta \in I\}. (**)
    2. y \in J \Rightarrow y \in I.
    3. x \in A_\beta \text{ for some } \beta \in J \Rightarrow x \in A_\beta \text{ for some } \beta \in I.
    4. \bigcup_{\alpha \in J} A_\alpha \subset \bigcup_{\alpha \in I} A_\alpha. \; \Box

    EDIT: S = the universe.

    I just want to verify that this works, particularly the third step.
    Last edited by phys251; January 3rd 2014 at 04:38 PM.
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  2. #2
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    Re: Union of indexed sets

    Quote Originally Posted by phys251 View Post
    "\text{Given } \{A_\alpha\}_{\alpha \in I}, J\subset I, \text{show } \bigcup_{\alpha \in J} A_\alpha \subset\bigcup_{\alpha \in I} A_\alpha."
    Here is what I did:
    1. \bigcup_{\alpha \in J} A_\alpha = \{x \in S: x \in A_\beta \text{ for some } \beta \in J\}, (*)
    \bigcup_{\alpha \in I} A_\alpha = \{x \in S: x \in A_\beta \text{ for some } \beta \in I\}. (**)
    2. y \in J \Rightarrow y \in I.
    3. x \in A_\beta \text{ for some } \beta \in J \Rightarrow x \in A_\beta \text{ for some } \beta \in I.
    4. \bigcup_{\alpha \in J} A_\alpha \subset \bigcup_{\alpha \in I} A_\alpha. \; \Box
    Frankly, I would not accept this as a proof.
    Your step #1 is totally unnecessary. What is S~?

    The first is: suppose that x\in\bigcup_{\alpha \in J} A_\alpha~.

    Now you must show that x\in\bigcup_{\alpha \in I} A_\alpha~..

    Take your other steps and do that. It is not complicated.
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    Re: Union of indexed sets

    Quote Originally Posted by phys251 View Post
    "\text{Given } \{A_\alpha\}_{\alpha \in I}, J\subset I, \text{show } \bigcup_{\alpha \in J} A_\alpha \subset\bigcup_{\alpha \in I} A_\alpha."

    Here is what I did:
    1. \bigcup_{\alpha \in J} A_\alpha = \{x \in S: x \in A_\beta \text{ for some } \beta \in J\}, (*)
    \bigcup_{\alpha \in I} A_\alpha = \{x \in S: x \in A_\beta \text{ for some } \beta \in I\}. (**)
    2. y \in J \Rightarrow y \in I.
    3. x \in A_\beta \text{ for some } \beta \in J \Rightarrow x \in A_\beta \text{ for some } \beta \in I.
    4. \bigcup_{\alpha \in J} A_\alpha \subset \bigcup_{\alpha \in I} A_\alpha. \; \Box

    I just want to verify that this works, particularly the third step.
    maybe this is a stupid question but what is S?

    I think you are making this harder than it is. All you need to show is that

    x \in \displaystyle{\bigcup_{\alpha \in J}}A_{\alpha} \longrightarrow x \in \displaystyle{\bigcup_{\alpha \in I}}A_{\alpha}

    It's a one liner to show this. You've got most of it already, just clean it up a bit.

    <snap beaten to the tape again>
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    Re: Union of indexed sets

    S is the universe, the set of all sets. Sorry.
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    Re: Union of indexed sets

    Quote Originally Posted by phys251 View Post
    S is the universe, the set of all sets. Sorry.
    There is no set of all sets.

    To quote Paul Halmos, Nothing contains everything!
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    Re: Union of indexed sets

    Is it really as simple as...

    1. Suppose $x \in \bigcup_{\alpha \in J} A_\alpha$ and $J \subset I.$
    2. Then $x \in \bigcup_{\alpha \in I} A_\alpha.$

    ...that? I figured there would be more to it.
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    Re: Union of indexed sets

    Quote Originally Posted by phys251 View Post
    Is it really as simple as...

    1. Suppose $x \in \bigcup_{\alpha \in J} A_\alpha$ and $J \subset I.$
    2. Then $x \in \bigcup_{\alpha \in I} A_\alpha.$

    ...that? I figured there would be more to it.
    It's a bit more detailed than that. Show if x is in the union over J it must lie in at least one A_{\alpha} with \alpha in J. Then as J is a subset of I that \alpha must lie in I and thus x lies in A_{\alpha} with \alpha in I and thus x lies in the union over I.

    Now put all that into math.
    Thanks from phys251
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    Re: Union of indexed sets

    Quote Originally Posted by Plato View Post
    To quote Paul Halmos, Nothing contains everything!

    very zen
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    Re: Union of indexed sets

    Wait...is this it?

    1. Suppose $x \in \bigcup_{\alpha \in J} A_\alpha.$
    2. Then $x \in A_\beta $ for some $\beta \in J.$
    3. Since $J \subset I, x \in A_\beta $ for some $\beta \in I.$
    4. Then $x \in \bigcup_{\alpha \in I} A_\alpha.$

    If not, then I'm completely stuck.
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    Re: Union of indexed sets

    Quote Originally Posted by phys251 View Post
    Wait...is this it?

    1. Suppose $x \in \bigcup_{\alpha \in J} A_\alpha.$
    2. Then $x \in A_\beta $ for some $\beta \in J.$
    3. Since $J \subset I, x \in A_\beta $ for some $\beta \in I.$
    4. Then $x \in \bigcup_{\alpha \in I} A_\alpha.$

    If not, then I'm completely stuck.
    why bring \beta into it but otherwise I suppose that's good enough.

    enclose your union symbol and under text in \displaystyle{} to get that text to be under the union symbol.

    x \in \displaystyle{\bigcup_{\alpha \in I}} A_\alpha

    becomes

    x \in \displaystyle{\bigcup_{\alpha \in I}} A_\alpha
    Last edited by romsek; January 3rd 2014 at 05:35 PM.
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    Re: Union of indexed sets

    Honestly, I don't care about style. I care that the proof in post #9 works. Does it?
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    Re: Union of indexed sets

    Quote Originally Posted by phys251 View Post
    Honestly, I don't care about style. I care that the proof in post #9 works. Does it?
    I said it's good enough. I don't think you need to add the index \beta to the mix but the way you've used it is correct.

    I wasn't trying to judge you on style. I was just providing you with some info I thought you might find useful.
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    Re: Union of indexed sets

    Okay. Thanks!
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    Re: Union of indexed sets

    Quote Originally Posted by romsek View Post
    very zen
    Indeed, the empty set's members have EVERY property, even mutually exclusive ones (for example, it is a subset of both of two disjoint sets, and lies in the intersection of even and odd integers, for example). Indeed, the essence of a set lies in *distinction* of otherwise formless form: as soon as we say: "x is in S", we have the notion of "inside" (membership) and "outside" (exclusion), a duality which in classical logic is borne by the idea of "not" (complementation induces disjoint union). This idea is turned into a (somewhat) elegant formal propositional arithmetic in a charming book "The Laws Of Form" (whose author escapes me at the moment).

    In other words, "somethingness" is something LESS than "nothingness", for when we say it is this or that, we are diminishing the limitless possibility of it being "anything" (even an impossible thing like triple-headed pink flying elephants). I believe this seeming paradox lies at the heart of all linguistic systems (including, for example, English in everyday discourse), where what we say is always somewhat less than what actually is, we filter to give meaning (or context) (this is the purpose of the "universal set of discourse", S: to "set the stage for future reference"...it is a bit distressing that an "all-encompassing context" (at least for "enough" mathematics) does not, properly speaking, exist, but it's not a perfect world, is it?).

    On topic, I agree with romsek, there is no need to introduce a second dummy index, the original alpha will serve throughout:

    \displaystyle x \in \bigcup_{\alpha \in J} A_{\alpha} \implies \exists \alpha \in J: x \in A_\alpha \implies \exists \alpha \in I: x \in A_{\alpha} \implies x \in \bigcup_{\alpha \in I} A_{\alpha}

    where the second implication is because J \subseteq I (the first and third implications are actually "two-way" (iff) although that is not necessary to note).

    To a certain extent, this is a minor quibble, but the point is: in mathematics, one often strives for CLARITY, so keeping "variables" to a minimum helps avoid confusion.
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