Check my work on these proofs? (basic set theory)

[Ryanodie]

Homework Statement

[/B]
I am going through Apostol's Calculus volume 1 and am working through I 2.5 #3. I'm not very familiar with doing proofs so I just wanted to make sure that I got the right idea here.

Here's the question:

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

Prove:
1. $A \subset B$
2. $A \subseteq B$
3. $1 \in A$

Homework Equations

Basic set notation

The Attempt at a Solution

1.
A is said to be a proper subset of B if $x \in A$ is in B and if $B \ne A$.
$x \in B$.
$B \ne A$.
Therefore, A is a proper subset of B.

2.
A is said to be a subset of B if every element $x \in A$ is also in B.
$x \in B$
Therefore, A is a subset of B.

3.
x is said to be an element of A if $x \in A$
x=1
$x \in A$
Therefore, $1 \in A$

I don't really know how to compose these, so this was my best shot. Let me know if I did alright or if their is a different way to write these (or if it's flat out wrong).

 

[micromass]

1.
A is said to be a proper subset of B if $x \in A$ is in B and if $B \ne A$.
$x \in B$.
$B \ne A$.
Therefore, A is a proper subset of B.

2.
A is said to be a subset of B if every element $x \in A$ is also in B.
$x \in B$
Therefore, A is a subset of B.

I cannot count these as correct. Maybe it's because the result is so obvious, that it is difficult to write a correct proof for it. Let us start with $A\subseteq B$. You have to show that each element of $A$ is also in $B$. So it suffices to take each element from $A$ and to see whether it is in $B$. So what are the elements of $A$? Are they in $B$?

x is said to be an element of A if $x \in A$
x=1
$x \in A$
Therefore, $1 \in A$

Yes, but it is written awkwardly. A proof must consist of full sentences and not many symbols. In this case, the result is so obvious that it doesn't really deserve a proof...

 

[Ryanodie]

I cannot count these as correct. Maybe it's because the result is so obvious, that it is difficult to write a correct proof for it. Let us start with $A\subseteq B$. You have to show that each element of $A$ is also in $B$. So it suffices to take each element from $A$ and to see whether it is in $B$. So what are the elements of $A$? Are they in $B$?
Yes, but it is written awkwardly. A proof must consist of full sentences and not many symbols. In this case, the result is so obvious that it doesn't really deserve a proof...

That's exactly what is throwing me. It's such a simple conclusion that I just don't know how to write more than what was originally stated.

 

[Keen94]

That's exactly what is throwing me. It's such a simple conclusion that I just don't know how to write more than what was originally stated.

Google: The Book of Proof by Richard Hammack or type in this link http://www.people.vcu.edu/~rhammack/BookOfProof/
Go to Part III section 8. You should be able to find your way a little better.

 

[Fredrik]

1.
A is said to be a proper subset of B if $x \in A$ is in B and if $B \ne A$.
$x \in B$.
$B \ne A$

You need to write the proof as a sequence of statements that the reader can verify are true. The problem with your statements is that they're not actually statements. Consider these examples:

Define x=1. We have $x^2=1$.
Define x=2. We have $x^2=1$.
For all real numbers x, $x^2=1$
There exists a real number x, such that $x^2=1$.
$x^2=1$

Each of the first four lines says something that's either true or false, but the fifth line doesn't (if x hasn't previously been assigned a value). Unfortunately some of your statements are of the same type as the last line. They're neither true nor false.

To avoid this, you need to make sure that every variable that you haven't assigned a value is the target of a "for all" or a "there exists".

In problem 3, there is literally nothing to prove. The only meaningful way to answer is to show that you know how to say $1\in A$ and $A=\{1,2\}$ in plain English.

 

[HallsofIvy]

To show that "A\subset B" you must show that "if x is in A then x is in B". But you can't just say "x is in B", you have to use the properties and definitions of A and B to show this is true. Here, "if x is in A" then x= 1 because the only member of A is 1. And B is defined as {1, 2} so clearly 1 is in B. Therefore A is a subset of B. To show it is a proper subset we only have to observe that 2 in B but not in A.

 

[Ryanodie]

Ok! Given the feedback and taking a look at the PDF file that was linked, I am going to try and give it another go with number 1. If it goes well, I'll try number 2.

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

Prove:
1. A⊂B
__________________________

1.
A is said to be a proper subset of B if $x \in A$ is in B and $A \ne B$.
Suppose that x = 1.
This shows that x is in A and also in B.
Now suppose that x=2.
In this case, x is in B but not in A. This shows that $A \ne B$.
We can now see that A is a proper subset of B.

 

[HallsofIvy]

Ok! Given the feedback and taking a look at the PDF file that was linked, I am going to try and give it another go with number 1. If it goes well, I'll try number 2.

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

Prove:
1. A⊂B
__________________________

1.
A is said to be a proper subset of B if $x \in A$ is in B and $A \ne B$.
Suppose that x = 1.
This shows that x is in A and also in B.
Now suppose that x=2.
In this case, x is in B but not in A. This shows that $A \ne B$.
We can now see that A is a proper subset of B.

This is still not sufficient. You started by saying "Suppose that x= 1" and observed that it is in A but not in B. But you did not use the fact that 1 is the only member of A. From what you said, it is quite possible that exist some other member of A that is not in B.

In my previous post, I started "If x is in A" and used the fact that A={1} to observe that x must be equal to 1. Think about the difference between that and what you said.

 

[Fredrik]

A is said to be a proper subset of B if $x \in A$ is in B and $A \ne B$.

This statement raises the question of what x is. In this statement, it should be the target of a "for all".

Suppose that x = 1.
This shows that x is in A and also in B.

No need to involve a variable. The statement "1 is an element of both A and B" accomplishes the same thing.

Now suppose that x=2.
In this case, x is in B but not in A. This shows that $A \ne B$.

It's sufficient to say that 2 is an element of B, and not an element of A.

You're stating the definition of the notation, and then you're saying that 1 is an element of A and B, and that 2 is an element of B but not A. This is adequate, I think, but I would consider it an improvement if you made it more clear that you're working through the list of elements of A one at a time, to make sure that each element of A is an element of B. You could also refer to the definitions of A and B. For example, you could say that by definition of A, 1 is the only element of A, and by definition of B, 1 is an element of B. This proves that each element of A is an element of B.

 

[Ryanodie]

Why is it important to worry about other elements of A if we can show with just those two that A is both in B and also not equal because it doesn't contain one of B's elements?

Edit: I get it now. I'm saying the difference between what could be

$A \cap B$

instead of

$A \subset B$

 

[vela]

Why is it important to worry about other elements of A if we can show with just those two that A is both in B and also not equal because it doesn't contain one of B's elements?

Edit: I get it now. I'm saying the difference between what could be

$A \cap B$

instead of

$A \subset B$

I'm not sure what you mean here.

In your argument, you assumed $x=1$, then said $x \in A$ and $x \in B$. All you've really argued is that 1 is an element of both A and B. Note that what you wrote would hold for the sets A={1,3} and B={1,2}, but in that case, A is clearly not a subset of B.

To prove that a set A is a subset of set B, you need to show that for all $x$, $x\in A$ implies $x \in B$. In other words, you want to show is if $x \in A$, then $x \in B$. So start with the assumption that $x \in A$. Write "Let $x \in A$" first and go from there. What can you say about $x$ given what you know about the set A? Show step by step that you can conclude that $x \in B$.