# Thread: Union of indexed sets

1. ## 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.

2. ## Re: Union of indexed sets

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

3. ## Re: Union of indexed sets

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

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>

4. ## Re: Union of indexed sets

S is the universe, the set of all sets. Sorry.

5. ## Re: Union of indexed sets

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

To quote Paul Halmos, Nothing contains everything!

6. ## 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.

7. ## Re: Union of indexed sets

Originally Posted by phys251
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.

8. ## Re: Union of indexed sets

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

very zen

9. ## 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.

10. ## Re: Union of indexed sets

Originally Posted by phys251
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

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

11. ## Re: Union of indexed sets

Honestly, I don't care about style. I care that the proof in post #9 works. Does it?

12. ## Re: Union of indexed sets

Originally Posted by phys251
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.

13. ## Re: Union of indexed sets

Okay. Thanks!

14. ## Re: Union of indexed sets

Originally Posted by romsek
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.