# Alternative proof to a trivial problem

## Main Question or Discussion Point

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.

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PeroK
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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.

Math_QED
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 An={x∈E: d(x,Ac)>1/n}, n∈ ℕ.Prove that i)A=∪n∈ℕ An

PeroK
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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 An={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.

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=Ao, 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 (Ac)c=A.What should I do?Try with the definitions, look for some other theorems?Any help provided is good.Thank you.

PeroK
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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=Ao, 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 (Ac)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##.

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

PeroK
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Thank for the answer PeroK,The steps you said I did try them but I could not continue. d(x,Ac):= inf{d(x,z): z∈Ac} if that helps.
That gets you started. What about showing that ##\bigcup A_n \subseteq A##?

That should not be too hard.

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

PeroK
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I can not combine them.A little hint if you can.I think if z∈∪An then d(z,Ac)>1 or d(z,Ac)>1/2 or...or d(z,Ac)>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##?

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

We need to show that after z∈An, then z∈Ao. We do something with the metrics I suppose.

PeroK
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If A⊆Ao, then if z∈A⇒z∈Ao .This is for A being an open set. It think that if d(x,Ac)>1/n then d(x,Ac) 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.

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

PeroK
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It has two other subexercises, the one I have solved it, but the other I have not.It says:iii) for every n∈ℕ, An is open set.
Okay, but to be honest you haven't shown any progress on this problem yet.

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

PeroK
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Perhaps if d(z,Ac)>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?

Last edited:
universe function
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
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##d(z, A^c) > \frac 1 n \ \Rightarrow \ z \notin A^c##.
Do you see why this is true?

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
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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?

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

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

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
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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##