flyingpig said:
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.
