Alternative proof to a trivial problem

[trees and plants]

Hi.I have this trivial problem for a metric d(x,y) that d((x,y)≥0. My alternative proof is 2d(x,y)=√4d²(x,y)=√d²(x,y)+d²(y,x)+2d(x,y)d(y,x)=√(d(x,y)+d(y,x))²≥d(x,x)=0 .Well it perhaps is a trivial proof but I did not know of this proof so I wanted to post it. Do you know other alternative proofs of this or other elementary or not so trivial problems in topology of metric spaces?The book I read had other proof for this problem.

 

[PeroK]

Hi.I have this trivial problem for a metric d(x,y) that d((x,y)≥0. My alternative proof is 2d(x,y)=√4d²(x,y)=√d²(x,y)+d²(y,x)+2d(x,y)d(y,x)=√(d(x,y)+d(y,x))²≥d(x,x)=0 .Well it perhaps is a trivial proof but I did not know of this proof so I wanted to post it. Do you know other alternative proofs of this or other elementary or not so trivial problems in topology of metric spaces?The book I read had other proof for this problem.

Unless $d(x, y) \ge 0$, then $d(x, y) = \sqrt{d(x, y)^2}$ does not hold.

 

[trees and plants]

You are correct.I made a mistake.I will post another exercise.Let us have A a non empty, open and subset of a metric space E.Let us also have A_n={x∈E: d(x,A^c)>1/n}, n∈ ℕ.Prove that i)A=∪_n∈ℕ A_n

 

[PeroK]

You are correct.I made a mistake.I will post another exercise.Let us have A a non empty, open and subset of a metric space E.Let us also have A_n={x∈E: d(x,A
^c)>1/n}, n∈ ℕ.Prove that i)A=∪_n∈ℕ A
_n

You should try some Latex:

https://www.physicsforums.com/help/latexhelp/

In any case, we need to see your best attempt at this problem.

 

[trees and plants]

ok, i will tell what I have figured out.We know that a metric space can be represented as a union of spherical neighborhoods with the same center. By definition an open set is one that A=A^o, and the interior of A containts the interior points of A, which means that if y is an element of A, then for a metric d and a neighborhood B(y,r), we have that B(y,r)⊆A, where B(y,r)={x∈E: d(x,y)<r}.Also (A^c)^c=A.What should I do?Try with the definitions, look for some other theorems?Any help provided is good.Thank you.

 

[PeroK]

ok, i will tell what I have figured out.We know that a metric space can be represented as a union of spherical neighborhoods with the same center. by definition an open set is one that A=A^o, and the interior of A containts the interior points of A, which means that if y is an element of A, then for a metric d and a neighborhood B(y,r), we have that B(y,r)⊆A, where B(y,r)={x∈E: d(x,y)<r}.Also (A^c)^c=A.What should I do?Try with the definitions, look for some other theorems?Any help provided is good.Thank you.

If you have to prove that two sets are equal, then one approach is to show that each is a subset of the other. In this case you need to show that:

1) $A \subseteq \bigcup A_n$; and,

2) $\bigcup A_n \subseteq A$

To do this, you could show that:

1) $x \in A \ \Rightarrow \ x \in \bigcup A_n$; and,

2) $x \in \bigcup A_n \ \Rightarrow \ x \in A$.

 

[trees and plants]

Thank for the answer PeroK,The steps you said I did try them but I could not continue. d(x,A^c):= inf{d(x,z): z∈A^c} if that helps.

 

[PeroK]

Thank for the answer PeroK,The steps you said I did try them but I could not continue. d(x,A^c):= inf{d(x,z): z∈A^c} if that helps.

That gets you started. What about showing that $\bigcup A_n \subseteq A$?

That should not be too hard.

 

[trees and plants]

I can not combine them.A little hint if you can.I think if z∈∪A_n then d(z,A^c)>1 or d(z,A^c)>1/2 or...or d(z,A^c)>1/n.Do I need to do something with the infimum?

 

[PeroK]

I can not combine them.A little hint if you can.I think if z∈∪A_n then d(z,A^c)>1 or d(z,A^c)>1/2 or...or d(z,A^c)>1/n.

Okay, a neater way to say that is that if $z \in \bigcup A_n$, then $\exists n: \ z \in A_n$.

Can you show that $A_n \subseteq A$?

 

[trees and plants]

If A⊆A^o, then if z∈A⇒z∈A^o .This is for A being an open set. It think that if d(x,A^c)>1/n then d(x,A^c) is a supremum for the set B={1/n: n∈ℕ}.

 

[trees and plants]

We need to show that after z∈A_n, then z∈A^o. We do something with the metrics I suppose.

 

[PeroK]

If A⊆A^o, then if z∈A⇒z∈A^o .This is for A being an open set. It think that if d(x,A^c)>1/n then d(x,A^c) is a supremum for the set B={1/n: n∈ℕ}.

This is getting muddled. Let me show you a simple proof of something else that might help.

Suppose we have a sequence of sets $S_n$ and $\forall n: S_n \subseteq S$. Then we can show that:
$$\bigcup S_n \subseteq S$$.

Proof: Let $x \in \ \bigcup S_n$. Then $\exists n: x \in S_n$. But, $S_n \subseteq S$, hence $x \in S$.

We have shown that:
$$x \in \ \bigcup S_n \ \Rightarrow x \in S$$
Which is equivalent to:
$$\bigcup S_n \subseteq S$$
Now, that result may help you a little. But, more important, you need to try to learn that style of logical proof writing.

 

[trees and plants]

It has two other subexercises, the one I have solved it, but the other I have not.It says:iii) for every n∈ℕ, A_n is open set.

 

[PeroK]

It has two other subexercises, the one I have solved it, but the other I have not.It says:iii) for every n∈ℕ, A_n is open set.

Okay, but to be honest you haven't shown any progress on this problem yet.

 

[trees and plants]

Perhaps if d(z,A^c)>1/n⇒d(z,A)≤1/n ? I hope that one helps.

 

[PeroK]

Perhaps if d(z,A^c)>1/n⇒d(z,A)≤1/n ? I hope that one helps.

What about $d(z, A^c) > \frac 1 n \ \Rightarrow \ z \notin A^c$. Is that even better?

 

[trees and plants]

It is almost unbelievable how I could not think of it.Sorry for that.It follows after that, but my question is that we are talking about the infimum,not the metric of two points.

 

[PeroK]

$d(z, A^c) > \frac 1 n \ \Rightarrow \ z \notin A^c$.

Do you see why this is true?

 

[trees and plants]

I suppose that if it is the infimum then every other metric≥ satisfies the condition. I have a little flaw with doing mistakes in proofs.How do we know when a proof is fully correct?From experience perhaps?

 

[PeroK]

I suppose that if it is the infimum then every other metric≥ satisfies the condition.

That's not the point. The point is that $z \in A^c \ \Rightarrow \ d(z, A^c) = 0$.

Do you undertstand the idea of contraposition?

 

[trees and plants]

You are right, incredible what applications theorems have.I now see why it is true.Perhaps I need to reconsider the theorems I know.

 

[trees and plants]

It is that I take the wrong ways to follow, something like I think I have prejudices about math.

 

[trees and plants]

That's not the point. The point is that $z \in A^c \ \Rightarrow \ d(z, A^c) = 0$.

Do you undertstand the idea of contraposition?

Yes I do.(p⇒q) ⇒ (~q⇒~p ) where p, q are propositions, in propositional logic.

 

[PeroK]

Yes I do.(p⇒q) ⇒ (~q⇒~p ) where p, q are propositions, in propositional logic.

So you understand in this case that:

$z \in A^c \ \Rightarrow \ d(z, A^c) = 0$.

Is the same as:

$d(z, A^c) > 0 \ \Rightarrow \ z \notin A^c \ \Rightarrow z \in A$

 

[trees and plants]

So you understand in this case that:

$z \in A^c \ \Rightarrow \ d(z, A^c) = 0$.

Is the same as:

$d(z, A^c) > 0 \ \Rightarrow \ z \notin A^c \ \Rightarrow z \in A$

Yes I do understand.How about the converse of it and other question that for every n∈ℕ, A_n is open set ?Thank you .

 

[PeroK]

Yes I do understand.How about the converse of it and other question that for every n∈ℕ, A_n is open set ?Thank you .

You have to make some effort on the converse.

 

[trees and plants]

We know that A⊆A ^- and d(z,A)=d(z,A^-), (A^c)^-=(A^o)^c and A⊆A^o for A as an open set, if A⊆B⇒B^c⊆A^c.How should I proceed?

 

[PeroK]

We know that A⊆A ^- and d(z,A)=d(z,A^-), (A^c)^-=(A^o)^c and A⊆A^o for A as an open set, if A⊆B⇒B^c⊆A^c.How should I proceed?

I don't see the relevance of $A^-$.

Let $x \in A \dots$. And remember that $A$ is open.

See post #6.

 

[trees and plants]

I think if A⊆A_n then A_n⊆∪_n∈ℕA_n. I do not know what to do next.I tried somehow.

 

[PeroK]

I think if A⊆A_n then A_n⊆∪_n∈ℕA_n. I do not know what to do next.I tried somehow.

That can't be right. We already know that every $A_n$ is a subset of $A$. What you need to do is to show that every element of $A$ is in at least one of the $A_n$.

Note that $A$ is open and, therefore, every $x \in A$ has a neighbourhood in $A$. That's the defining property of an open set. That geometrically is very close to what you need to show. If we take $x \in A$ then for some $n$ we have $x \in A_n$.

You need to complete the proof rigorously, but do you see the idea?

 

[trees and plants]

I thought of something, but I suppose it is wrong. I will say it: A is open, so we have B(x,r)⊆A and for y∈B(x,r) also y∈A. d(x,y)<r, so d(x,y)=r/2 and because d(x,A^c )>1 we choose a y for d(y,A^c)=1+1/2 and we take as r=10/2,so d(x,A^c)≥d(x,y)-d(A^c,y)=r/2-3/2=10/4-3/2=1.

 

[PeroK]

I thought of something, but I suppose it is wrong. I will say it: A is open, so we have B(x,r)⊆A and for z∈B(x,r) also z∈A. d(x,y)<r, so d(x,y)=r/2 and because d(x,A^c )>1 we choose a y for d(y,A^c)=1+1/2 and we take as r=10/2,so d(x,A^c)≥d(x,y)-d(A^c,y)=r/2-3/2=10/4-3/2=1.

I can't follow any of that. You must get into the habit of specifying what things are and using the quantifiers $\forall$ and $\exists$. What are $x, y, z, r$? And why is $d(x, A^c) > 1$?

To give you a bit more help, you could start with:

Let $x \in A$. As $A$ is open, there exists a neighbourhood of $x$ within $A$. I.e. $\exists r: B(x, r) \subseteq A$.

 

[trees and plants]

If we took the union of B(x,r_x) and the union of A_n?Would it help?Because a set is open if it can be represented as a union of spherical neighborhoods.

 

[PeroK]

If we took the union of B(x,r_x) and the union of A_n?Would it help?

Which $n$? To do what? You're trying to find $n$ where $x \in A_n$.

Hint: $n$ must be related to $r$ somehow.