Re: Topology beginer some help with proofs please....

Quote:

Originally Posted by

**Bwts** If $\displaystyle B \subset \bigcap_{\alpha \in I}$ then for every $\displaystyle x \in B$, $\displaystyle x \in \bigcap_{\alpha \in I} A_\alpha$

If $\displaystyle x \in \bigcap_{\alpha \in I} A_\alpha$ then for every $\displaystyle \beta \in I$, $\displaystyle x \in A_\beta$

That's correct. Let's call these statements (1) and (2).

Quote:

Originally Posted by

**Bwts** So $\displaystyle x \in B$ if and only if $\displaystyle B \subset A_\beta$

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 $\displaystyle B\subset A_\beta$ for all $\displaystyle \beta$. However, x can be arbitrary; it can belong to B or not.

Combine (1) and (2) and write $\displaystyle B \subset \bigcap_{\alpha \in I} A_\alpha$ without using $\displaystyle \subset$ or $\displaystyle \bigcap$. Then do the same with the right-hand side, i.e., "for each $\displaystyle \beta \in I$, $\displaystyle B \subset A_\alpha$." Compare the results.

Re: Topology beginer some help with proofs please....

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

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

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

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

R. H. S.

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

Are (3) & (4) equivalent (they look like theyare to me but the indices are throwing me)?

Re: Topology beginer some help with proofs please....

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

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

or:

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

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 ⊂ A_{1}) & (B ⊂ A_{2})] → (B ⊂ (A_{1}∩A_{2}))

the "&" is most important, here.

Re: Topology beginer some help with proofs please....

Quote:

Originally Posted by

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

Should be $\displaystyle A_\beta $.

Quote:

Originally Posted by

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

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

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

Yes, but I would replace "If... then..." in (1) and (2) by "iff."

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Originally Posted by

**Bwts** R. H. S.

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

Should be $\displaystyle A_\beta $.

Quote:

Originally Posted by

**Bwts** 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.

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

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

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

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

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

Doesnt look right to me, is it?

Re: Topology beginer some help with proofs please....

Quote:

Originally Posted by

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

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

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

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

(1) and (2) are correct.

Quote:

Originally Posted by

**Bwts** So $\displaystyle \bigcup_{\alpha \in I} A_\alpha \subset B$ iff for each $\displaystyle \beta \in I$, $\displaystyle A_\beta \subset B$ (as we dont know which $\displaystyle A_\beta$ 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 $\displaystyle \beta\in I$ such that $\displaystyle x\in A_\beta$, then $\displaystyle x\in B$.

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

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

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

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

It is important that in (3), the whole $\displaystyle \exists\beta\in I.\,x\in A_\beta$ is in the premise of the implication. Now, one of logical laws says that $\displaystyle (\exists x\,P(x))\to Q$ is equivalent to $\displaystyle \forall x.\,(P(x)\to Q)$, 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 $\displaystyle \forall$ to get (4).

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

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

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

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

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

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

thus $\displaystyle x \in B$, since for any $\displaystyle \beta \in I,\ A_\beta \subset B$, which shows that $\displaystyle \bigcup_{\alpha \in I} A_\alpha \subset B$.

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.

:)

Re: Topology beginer some help with proofs please....

OK I'm really struggling here...

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

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

My attempt...

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

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

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

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

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

Attempt one...

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

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

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

Attempt two...

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

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

....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.

Re: Topology beginer some help with proofs please....

Quote:

Originally Posted by

**Bwts**

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

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

..............

Attempt two...

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

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

So does this directly imply that $\displaystyle \bigcup_{\beta \in I} A_\beta \subset \bigcup_{\beta \in I} B_\beta$?

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

Just because $\displaystyle \beta \in I$ and $\displaystyle \alpha \in I$ doesnt make $\displaystyle \beta = \alpha$ does it? What am I missing here?

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

My attempt...

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

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

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

It's useless to say "If $\displaystyle x \in \bigcup_{\alpha \in I} A_\alpha$." 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. Quote:

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.