# Prove that the set of points on any two line segments are equivalent.

• Mar 4th 2010, 04:31 PM
davismj
Prove that the set of points on any two line segments are equivalent.
http://i46.tinypic.com/2n1cinl.jpg

I'm guessing this is what the question is asking?

I think its pretty show that there is an equivalence relation here. And then the implication would be that since you can take a line segment inside another line segment and come up with an equivalence relation, the set of points on any given line segment is infinite.

Right?
• Mar 4th 2010, 04:46 PM
Drexel28
Quote:

Originally Posted by davismj
http://i46.tinypic.com/2n1cinl.jpg

I'm guessing this is what the question is asking?

I think its pretty show that there is an equivalence relation here. And then the implication would be that since you can take a line segment inside another line segment and come up with an equivalence relation, the set of points on any given line segment is infinite.

Right?

Let $\displaystyle \ell_1,\ell_2$ be the two lines. Translate $\displaystyle \ell_1$ over however many units so that it is not above $\displaystyle \ell_2$. Consider the projections $\displaystyle \pi:\ell_1,\ell_2\mapsto(a,b),(c,d)$. This is clearly a bijection. Thus, we must merely show that given two segments $\displaystyle (a,b),(c,d)$ that $\displaystyle (a,b)\simeq (c,d)$. Conjecture there exists a linear bijection between them. Then $\displaystyle f(x)=mx+k$. So then we need that $\displaystyle {f(a)=ma+k=d}\brace{f(b)=mb+k=d}$. Solving this system of equations will give you a bijective function between the two which is actually a homeomorphism (that last part isn't necessary)

And, I don't see how defining an equivalence relation helps.

Wait....I read the question. But now I am having second thoughts about it's meaning. I thought you meant to show that any two lines has the same number of points("equivalent" in the category of sets). Is there somethign else?
• Mar 4th 2010, 04:55 PM
davismj
Quote:

Originally Posted by Drexel28
Let $\displaystyle \ell_1,\ell_2$ be the two lines. Translate $\displaystyle \ell_1$ over however many units so that it is not above $\displaystyle \ell_2$. Consider the projections $\displaystyle \pi:\ell_1,\ell_2\mapsto\mathbb{R}$. This is clearly a bijection. Thus, we must merely show that given two segments $\displaystyle (a,b),(c,d)$ that $\displaystyle (a,b)\simeq (c,d)$. Conjecture there exists a linear bijection between them. Then $\displaystyle f(x)=mx+k$. So then we need that $\displaystyle {f(a)=ma+k=d}\brace{f(b)=mb+k=d}$. Solving this system of equations will give you a bijective function between the two which is actually a homeomorphism (that last part isn't necessary)

And, I don't see how defining an equivalence relation helps.

Wait....I read the question. But now I am having second thoughts about it's meaning. I thought you meant to show that any two lines has the same number of lines ("equivalent" in the category of sets). Is there somethign else?

No, that's it for this question. To my understanding, the problem is to show that for any two arbitrary lines p and q in a plane, there is a bijection between P = {x : x is on p} and Q = {y : y is on q}.

That's why I parametrized each line. Each line is a bijection, from [0,1] to P and [0,1] to Q. Therefore, we can show that P ~ [0,1] ~ Q so P ~ Q. What do you think?

"First, we recall that set A is equivalent to set B if and only if there is a one-to-one mapping of A onto B."
• Mar 4th 2010, 04:58 PM
Drexel28
Quote:

Originally Posted by davismj
No, that's it for this question. To my understanding, the problem is to show that for any two arbitrary lines p and q in a plane, there is a bijection between P = {x : x is on p} and Q = {y : y is on q}.

That's why I parametrized each line. Each line is a bijection, from [0,1] to P and [0,1] to Q. Therefore, we can show that P ~ [0,1] ~ Q so P ~ Q. What do you think?

"First, we recall that set A is equivalent to set B if and only if there is a one-to-one mapping of A onto B."

Oh maybe...write out the bijection and I'll check it.

(what I said works then btw)
• Mar 4th 2010, 05:08 PM
davismj
Quote:

Originally Posted by Drexel28
Oh maybe...write out the bijection and I'll check it.

(what I said works then btw)

$\displaystyle F: f(p) = q_0 + \frac{q_0 - q_1}{p_0 - p_1}(p - p_0) = q$

and

$\displaystyle G: g(q) = p_0 + \frac{p_0 - p_1}{q_0 - q_1}(q - q_0) = p$
• Mar 4th 2010, 05:11 PM
Drexel28
Quote:

Originally Posted by davismj
$\displaystyle F: f(p) = q_0 + \frac{q_0 - q_1}{p_0 - p_1}(p - p_0) = q$

and

$\displaystyle G: f(q) = p_0 + \frac{p_0 - p_1}{q_0 - q_1}(q - q_0) = p$

That's a mapping between...
• Mar 4th 2010, 05:25 PM
davismj
Quote:

Originally Posted by Drexel28
That's a mapping between...

f(g(q)) = q

g(f(p)) = p

These statements imply that these functions are bijective, since both f(p) and g(q) have inverses.
• Mar 4th 2010, 05:29 PM
Drexel28
Quote:

Originally Posted by davismj
f(g(q)) = q

g(f(p)) = p

These statements imply that these functions are bijective, since both f(p) and g(q) have inverses.

Yes. But to have a mapping you need to have a rule of assignment, a domain, and a codmain. You've only given the rules of assignment. But I get what you're saying