Alternative proof to a trivial problem

[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.

 

[trees and plants]

If we consider r=1/n and y∈A^c?What is the solution?I do not know.

 

[PeroK]

If we consider r=1/n and y∈A^c?What is the solution?I do not know.

That's getting close. Technically, you let $n > 1/r$. Then you show that $\forall y \in A^c$, we have $d(x, y) > 1/n$.

That tells you that $d(x, A^c) \ge \frac 1 n > \frac 1 {n+1}$ and you're done.

You have to put that all together though.

 

[trees and plants]

I reached so far at least.It is just the details I had to deal with, although they are important and they are needed.Thank you.I must leave from the forum now.I will see you tomorrow hopefully.

 

[mathman]

I always thought that $d(x,y)\ge 0$ was part of the definition?

 

[PeroK]

I always thought that $d(x,y)\ge 0$ was part of the definition?

So did I, but apparently this condition can be deduced from the others.

 

[mathman]

So did I, but apparently this condition can be deduced from the others.

There seems to be a long series of posts which appear to lead to this conclusion. Could you summarize it, starting from the beginning?

 

[PeroK]

There seems to be a long series of posts which appear to lead to this conclusion. Could you summarize it, starting from the beginning?

From post #3 or #4 we switched to a new problem.

 

[mathman]

From post #3 or #4 we switched to a new problem.

If I understand you correctly, you never proved $d(x,y)\ge 0$ can be deduced, as opposed to being part of the definition.

 

[PeroK]

If I understand you correctly, you never proved $d(x,y)\ge 0$ can be deduced, as opposed to being part of the definition.

Not on this thread. It's on the wikipedia page if you are interested.