I am probably going to be killed by this course

[flyingpig]

Homework Statement

It was back in April where I asked the professor teaching if it was possible to take this course called "Analysis" without any proof-based math background and he answered with a "yes".

Now I am starting to regret a bit of it since we just started and i have so much to catch up. I wonder if he said "yes" just so he could fail me.

Anyways here is the problem I encountered

The set T = {0, {1,2,3},4,5}, there are four elements, why is $$2 \notin T$$?

If I call J = {1,2,3}, then T = {0, J,4,5}

Clearly it looks like $$J\in T$$ and $$2\in J$$

So why can't it be

$$2\in J \in T$$?

 

[Dick]

Because it's not true that if A is an element of B and B is an element of C then A is an element of C. It doesn't work that way. Look, you agree T has four elements, right? Is 2 one of them?

 

[flyingpig]

Because it's not true that if A is an element of B and B is an element of C then A is an element of C. It doesn't work that way. Look, you agree T has four elements, right? Is 2 one of them?

Ohh is that how we see it? So the "2 (or the 1 and 3 too!)" has been replaced by {...}

 

[Dick]

Ohh is that how we see it? So the "2 (or the 1 and 3 too!)" has been replaced by {...}

Yes, the set {1,2,3} is an element of T. 1, 2 and 3 aren't.

 

[Char. Limit]

$$2\in J \in T$$

This is true.

$$2 \in T$$

This is not. Even if A belongs to B, which belongs to C, that doesn't imply that A belongs to C. In other words...

$$A \in B \in C \not\rightarrow A \in C$$

 

[flyingpig]

What about

If I say

$$K = \left \{ \left. 2 \right \} \right.$$
$$T = \left \{ \left. 0, \left \{ \left. 1,2,3\right \} \right. ,4, 5 \right \} \right.$$
$$J = \left \{ \left. 1,K,3 \right \} \right.$$

It follows that $$K \subseteq J \subseteq T$$? DOes that work? If so why?? If not, then does it follow the above argument?

EDIT: does anyone have a good idea for replacing curvy brackets without \left \{ \left. \right \} \right.?

 

[Dick]

There is a big difference between 2 and {2}. The second is a set with one element, 2. The first one isn't a set at all.

 

[Char. Limit]

What about

If I say

$$K = \left \{ \left. 2 \right \} \right.$$
$$T = \left \{ \left. 0, \left \{ \left. 1,2,3\right \} \right. ,4, 5 \right \} \right.$$
$$J = \left \{ \left. 1,K,3 \right \} \right.$$

It follows that $$K \supseteq J \supseteq T$$? DOes that work? If so why?? If not, then does it follow the above argument?

EDIT: does anyone have a good idea for replacing curvy brackets without \left \{ \left. \right \} \right.?

Not quite. J in this case is the set $\left\{1, \{2\}, 3 \right\}$. That obviously doesn't belong to T.

 

[flyingpig]

I got the idea from my book

A set is called a subset of a set B if every element of A also belongs to B. If A is a subset of B, then we write A $$\subseteq$$ B. If A, B, and C are sets such that A $$\subseteq$$ B and B $$\subseteq$$ C, then A $$\subseteq$$ C.

 

[flyingpig]

Sorry I had to run and fix the curly brackets and made some typos instead...

$$K = \left \{ \left. 2 \right \} \right.$$
$$T = \left \{ \left. 0, \left \{ \left. 1,\left \{ \left. 2 \right \} \right.,3\right \} \right. ,4, 5 \right \} \right.$$
$$J = \left \{ \left. 1,K,3 \right \} \right.$$

 

[Dick]

I got the idea from my book

That's very true for "is a subset of". That property is called transitive property. "is an element of" is NOT transitive. They are two different things.

 

[flyingpig]

That's very true for "is a subset of". That property is called transitive property. "is an element of" is NOT transitive. They are two different things.

But they feel so familiar!

 

[flyingpig]

I might as well ask it here anyways, what does this mean?

R\Q

R = real number
Q = rational number

What does the \ mean?

 

[Char. Limit]

You would be dealing with subsets in a case like this:

J = {1, 2, 3}
T = {0, 1, 2, 3, 4, 5}

In this case, J is a subset of T, as every element of J is also an element of T. However, you get something like this:

J = {1, 2, 3}
T = {0, {1, 2, 3}, 4, 5}

In this case, J is an element of T, as J itself, not its elements, is an element of T. They seem similar but are very different.

 

[Char. Limit]

I might as well ask it here anyways, what does this mean?

R\Q

R = real number
Q = rational number

What does the \ mean?

A rough translation of that would be "excluding", so that R\Q would mean "The reals, excluding the rationals".

 

[Dick]

But they feel so familiar!

But 2 is NOT an element of {0, {1,2,3},4,5}. It's not one of the four elements. Period.

 

[flyingpig]

You would be dealing with subsets in a case like this:

J = {1, 2, 3}
T = {0, 1, 2, 3, 4, 5}

In this case, J is a subset of T, as every element of J is also an element of T. However, you get something like this:

J = {1, 2, 3}
T = {0, {1, 2, 3}, 4, 5}

In this case, J is an element of T, as J itself, not its elements, is an element of T. They seem similar but are very different.

Wow that clears up A LOT lolEDIT

So in other words (even though it does look like I am juist translating what you said)

In the case of

J = {1, 2, 3}
T = {0, {1, 2, 3}, 4, 5}

$$J \in T$$ and $$J \subseteq T$$

But 1, 2, 3 are NOT in T.

WHat happens if we change T to something like (and J stays as J = {1,2,3})

T = {1,{1,2,3},4,5}

Is it right to say

$$1 \in J \in T$$ or reorder it like $$1 \in T \in J$$?

A rough translation of that would be "excluding", so that R\Q would mean "The reals, excluding the rationals".

Good because the prof was writing this, I thought it meant something like "without".

Oh I am derailing my own thread, what would something like P /\ K mean? I think P and K are circle things - sets...

 

[Char. Limit]

P∧K means "P and K", at least in logic. In set theory, I believe $P \cap K$ refers to an intersection, or the set of all the elements that are elements of both P and K.

 

[SammyS]

What about

If I say

$$K = \left \{ \left. 2 \right \} \right.$$
$$T = \left \{ \left. 0, \left \{ \left. 1,2,3\right \} \right. ,4, 5 \right \} \right.$$
$$J = \left \{ \left. 1,K,3 \right \} \right.$$

It follows that $$K \subseteq J \subseteq T$$? DOes that work? If so why?? If not, then does it follow the above argument?

EDIT: does anyone have a good idea for replacing curvy brackets without \left \{ \left. \right \} \right.?

The set J here isn't the same as the set J of the O.P.

Here J = {1, K, 3} = {1, {2}, 3} .

So, $J\not\subseteq T\,.$

 

[process91]

So in other words (even though it does look like I am juist translating what you said)

In the case of

J = {1, 2, 3}
T = {0, {1, 2, 3}, 4, 5}

$$J \in T$$ and $$J \subseteq T$$

But 1, 2, 3 are NOT in T.

Let's address this first. This is, in fact, not true. The way to think about $A\subseteq B$ is that for each element that is in A, that element is in B. Think very literally, like a computer program here. Example:

$A=\left\{1,2,3\right\}$
$B=\left\{1,2,3,4,5,6,7,8\right\}$

I pick an element of A, say 1, and check to see if it is true that $1\in B$. Sure enough, it is. Now I pick another element of A. Order is unimportant, but it makes intuitive sense to pick 2. Again, I find that 2 is in B, or symbolically $2 \in B$. Finally we check the last element which is in A, 3, and find that $3 \in B$. Since every element which is in A is also in B, we have that A is a subset of B (because that's what the definition of subset is), so we could write $A \subseteq B$.

As an aside, don't get bogged down with the symbols, a lot of people like them because they are more concise, but a book I have on Real Analysis mentioned the fact that (especially if they are unfamiliar) it takes more time to read them because you have to untranslate and then retranslate every one back and forth into English. This book actually suggested avoiding symbolic notation whenever possible. Of course, if your course relies heavily on symbols it would be to your advantage to familiarize yourself with them.

Now, an example similar to yours:

$A = \left\{0,2,4 \right\}$
$B = \left\{0, \left\{0,2,4\right\}, 9, 7\right\}$

Here again we check an element of A, say 0, and see if it is in B. It is, so we have $0 \in B$. We're well on our way to proving that A is a subset of B. Now let's take 2, and see if 2 is in B. Well, B has four elements: 0, {0,2,4}, 9, and 7. None of these are the same thing as 2. One of these is a set which contains 2, but that's not nearly the same thing as 2, just as a box which contains an apple is not, itself, an apple.

So we tried, but could not find 2 in B. In fact, 2, is not in B, so A is not a subset of B.

It may help, in the beginning, to think of sets as boxes containing things. This is not a perfect analogy, as you will see when dealing with the empty set, however, it should serve adequately well for dealing with questions like these. Thinking this way, consider this:

A = {2}
B = {0,1,{2}}

Is A a subset of B, an element of B, or both? I would think about this in the following way:
1. Remember what the definition of a subset is, and go through the "computerized" process, keeping in mind that a box with an apple in it is not, itself, an apple.
2. Now, treating A as a single entity instead of a set, see if you can find A within B somewhere. To continue to use the box analogy, A is a box which contains one element, 2, so you're looking to see if B contains a box with only the number two in it.

Just like in real life, you can have boxes inside of boxes. This leads to some interesting paradoxes down the road, which you are sure to cover in class, and is partly why it is not a perfect analogy. Please remember that the box analogy is just a set of training wheels, which you will eventually have to shed.

 

[flyingpig]

OKay I just did that problem similar to that one and i am wrong

I got confused J is not a subset of T

 

[flyingpig]

A = {2}
B = {0,1,{2}}

A is NOT subset of B, but if I let K = {{2}}, then, it is

 

[process91]

I assume you mean that if you let K={{2}} then K is a subset of B. That is absolutely correct.

You are also correct that A is not a subset of B. Is A an element of B?

 

[flyingpig]

Yes A is in B. I am doing A LOT of things at once because I have so much to cover...

sorry if I am a little slow. This is what you get for believing your professor.

 

[flyingpig]

Yeah or just say K = {A}

 

[process91]

Nope, all good, I'm working on other stuff too.
I think we can probably flesh out all of the basic interesting stuff about sets pretty quickly, and expose any areas which might trip you up.

So, you've sort of addressed the second part of your post above,

Is it right to say

$$1 \in J \in T$$ or reorder it like $$1 \in T \in J$$?

In the example we just did with A and B, you had that A is in B, not B in A. Obviously 2 is in A, so we could write $2 \in A \in B$. It wouldn't make sense to change the order, since (as we mentioned) B is not in A. Again, thinking about them like boxes clears this up immediately.

As you can probably tell, I'm wary about leaving this boxes analogy out there for too long, because there are always some points which professors love to poke at which break the boxes analogy, so if you feel confident with this stuff I would just like to point out the empty set (did you cover that at all yet?) and how it breaks down the boxes analogy.

 

[HallsofIvy]

Yes A is in B. I am doing A LOT of things at once because I have so much to cover...

sorry if I am a little slow. This is what you get for believing your professor.

Perhaps you asked your professor the wrong question. You said, initially, that you asked if it were possible to do the course if you have never had an introduction to "proofs". That would give him/her no reason to think you knew nothing about sets. Most people are introduced to sets in a basic algebra or, at latest, precalculus course.

 

[flyingpig]

Perhaps you asked your professor the wrong question. You said, initially, that you asked if it were possible to do the course if you have never had an introduction to "proofs". That would give him/her no reason to think you knew nothing about sets. Most people are introduced to sets in a basic algebra or, at latest, precalculus course.

No here is the quote

Is it possible to take Math xyz without taking Math abc even though Math abc is a prerequsite?

And here was the laconic response

yes

 

[SammyS]

No here is the quote


And here was the laconic response

Maybe it's possible.

Perhaps it's not advisable.
...

fp,

Actually, I've given this some thought since I first saw Post #1 of this thread.

Having a fair amount of experience helping you in physics - mostly electrostatics - I have to admit that I'm really unsure of what advice to give you. Seeing the trouble you're having with some basic ideas about sets and basic properties of some elementary mathematical structures, I'm inclined to say that you probably need to take a more elementary proof-based course. However, you also had trouble understanding - or should I say accepting - some rather basic ideas in electrostatics, but in the end I assume you did alright in the end (or is that an unwarranted assumption on my part?)

I suspect that if you take a more elementary course than this Analysis course, you will still have similar issues to what you have been posting in this last week or so. However, I think it's more appropriate for you to sort out these issues at the more basic level.

Good Luck!

Either way I think there will be people here who will try to give you help.

 

[muzak]

http://www.uccs.edu/~math/vidarchive.html

I'd start with Discrete Mathematics in those archives, it introduces you to proofs and the basics. Registration is free.