Alternative proof to a trivial problem

[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.