Open sets in R being the union of open intervals

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SUMMARY

The discussion centers on the theorem regarding open sets in the real numbers R being the union of open intervals, specifically under the standard metric. It is established that while every metric equivalent to the standard metric maintains this property, the theorem does not extend to arbitrary metrics. The discrete metric is highlighted as a counterexample, where all sets, including singletons, are considered open, yet they do not represent unions of open intervals. The dependence of the openness of intervals on the chosen metric is emphasized as critical to the proof of the theorem.

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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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