Mathematical Reasoning and Writing - Counterexamples with subsets.

[mliuzzolino]

Homework Statement

Let f: A --> B be a function and let S, T \subseteq A and U, V \subseteq B.

Give a counterexample to the statement: If f (S) \subseteq f (T); then S \subseteq T:

Homework Equations

The Attempt at a Solution

PF:

Assume f(S) \subseteq f(T).

Let x \in S.

Then \exists y \in f(S) \ni f(x) = y.

Since f(S) \subseteq f(T), y \in f(T).

****

Suppose \forall a \in T where a ≠ x, \exists y \in f(T) \ni f(a) = y.

Then x \notin T.

Q.E.D.



I am not exactly sure I am doing this right, especially the reasoning beyond the ****. I almost have the feeling I should use the pre image of f(T) somehow to show that x \notin T.

Why can I not just say that \exists x \in S where x \notin T? Would that not suffice as a counterexample in such a general proof as this?

 

[Dick]

Homework Statement

Let f: A --> B be a function and let S, T \subseteq A and U, V \subseteq B.

Give a counterexample to the statement: If f (S) \subseteq f (T); then S \subseteq T:

Homework Equations

The Attempt at a Solution

PF:

Assume f(S) \subseteq f(T).

Let x \in S.

Then \exists y \in f(S) \ni f(x) = y.

Since f(S) \subseteq f(T), y \in f(T).

****

Suppose \forall a \in T where a ≠ x, \exists y \in f(T) \ni f(a) = y.

Then x \notin T.

Q.E.D.



I am not exactly sure I am doing this right, especially the reasoning beyond the ****. I almost have the feeling I should use the pre image of f(T) somehow to show that x \notin T.

Why can I not just say that \exists x \in S where x \notin T? Would that not suffice as a counterexample in such a general proof as this?

You aren't supposed to do a general proof. The statement isn't false for all functions, only some. You have to think of one.

 

[mliuzzolino]

You aren't supposed to do a general proof. The statement isn't false for all functions, only some. You have to think of one.

Oh! I don't know why I was thinking what I was.

How about...

Proof:
Let f: ℝ → [0, ∞) by f(x) = x².
Q.E.D.

The negative ℝ could be considered S and the positive ℝ could be considered T. Then by f(x) = x², f(S) is contained in f(T), but obviously S is not contained in T.

Would this be a suitable counterexample?

 

[micromass]

Oh! I don't know why I was thinking what I was.

How about...

Proof:
Let f: ℝ → [0, ∞) by f(x) = x².
Q.E.D.

The negative ℝ could be considered S and the positive ℝ could be considered T. Then by f(x) = x², f(S) is contained in f(T), but obviously S is not contained in T.

Would this be a suitable counterexample?

That's perfect!

 

[micromass]

Some comments on your OP. When writing a proof, you should never use the symbols $\exists$ and $\forall$. You should always write it out in words. This is a very common mistake that new people make and it's one way I see whether somebody is used to proving things or not.

And your proof should be much wordier. A proof should really be read like an english text.

 

[Curious3141]

Some comments on your OP. When writing a proof, you should never use the symbols $\exists$ and $forall$. You should always write it out in words. This is a very common mistake that new people make and it's one way I see whether somebody is used to proving things or not.

And your proof should be much wordier. A proof should really be read like an english text.

I had no idea walruses observe the 1st of April.

 

[micromass]

I had no idea walruses observe the 1st of April.

I was serious

 

[Curious3141]

I was serious

Haha, really?

Seriously, I do find a proof with all those logical operators a real PITA to read. I'd much prefer they wrote it all in words. But then, I'm not a mathematician, your walrus-ness.