Open sets in R being the union of open intervals

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The theorem stating that open sets in R can be expressed as the union of open intervals holds true for metrics equivalent to the standard metric, but does not generalize to arbitrary metrics. For instance, in the discrete metric, all sets are considered open, including singletons, which cannot be represented as unions of open intervals. Additionally, the concept of open intervals may not retain its properties under different metrics, affecting the proof's validity. The discussion highlights the importance of the metric's characteristics in determining the openness of sets. Understanding these nuances is essential for accurately applying the theorem in various contexts.
wisvuze
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Hello, I know one proof of this well known theorem that assumes on the metric of R being the standard metric. Does this result generalize to arbitrary metrics on R?
thank you
 
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Hi wisvuze! :smile:

Well, every metric equivalent to the standard metric satisfies this. But the result does not generalize to arbitrary metric spaces. For example, take the discrete metric

d(x,y)=1~\text{if}~x\neq y~\text{and}~d(x,x)=0

then all sets are open. In particular, the singletons are open. But the singletons are not the union of open intervals!

It can also happen that the open intervals are not open sets anymore!
 
Hi! Thanks :) I tried to prove it for arbitrary metrics but failed ( I wonder why now :P )
I should have realized though that the open-ness of the "open interval" depends on the metric ( in particular, one that acts something like the standard metric ); and the open ness of the intervals is the crucial part for any proof of this theorem that I know
 

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