# Homework Help: I am probably going to be killed by this course

1. Sep 9, 2011

### flyingpig

1. The problem statement, all variables and given/known data

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

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

2. Sep 9, 2011

### 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?

3. Sep 9, 2011

### flyingpig

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 has been replaced by {...}

4. Sep 9, 2011

### Dick

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

5. Sep 9, 2011

### Char. Limit

This is true.

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...

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

6. Sep 9, 2011

### 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.?

7. Sep 9, 2011

### 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.

8. Sep 9, 2011

### Char. Limit

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

9. Sep 9, 2011

### flyingpig

I got the idea from my book

10. Sep 9, 2011

### flyingpig

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

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

11. Sep 9, 2011

### Dick

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.

12. Sep 9, 2011

### flyingpig

But they feel so familiar!!

13. Sep 9, 2011

### 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?

14. Sep 9, 2011

### 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.

15. Sep 9, 2011

### Char. Limit

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

16. Sep 9, 2011

### Dick

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

17. Sep 9, 2011

### flyingpig

Wow that clears up A LOT lol

EDIT

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 $$1 \in T \in J$$?

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...

18. Sep 9, 2011

### 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.

19. Sep 9, 2011

### SammyS

Staff Emeritus
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, $J\not\subseteq T\,.$

Last edited: Sep 9, 2011
20. Sep 9, 2011

### process91

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.

Last edited: Sep 9, 2011
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