General topology: Prove a Set is Open

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SUMMARY

The set A = {x ∈ ℝ² : 1 < x² + y² < 2} is proven to be open in the context of general topology. The proof utilizes the properties of open balls and the triangle inequality to demonstrate that every point in A has a neighborhood contained within A. The complement of A is shown to be closed, confirming that A is not closed. The discussion emphasizes the importance of understanding the basis for the topology of ℝ², specifically the use of open balls for constructing open sets.

PREREQUISITES
  • Understanding of general topology concepts, specifically open and closed sets.
  • Familiarity with metric spaces and the properties of distance functions.
  • Knowledge of the triangle inequality in the context of Euclidean spaces.
  • Ability to work with open balls and their intersections in ℝ².
NEXT STEPS
  • Study the definition and properties of open sets in topology.
  • Learn about the triangle inequality and its applications in metric spaces.
  • Explore different bases for topologies, particularly in ℝ².
  • Investigate the relationship between open and closed sets in various topological spaces.
USEFUL FOR

Students of mathematics, particularly those studying topology, metric spaces, and analysis, will benefit from this discussion. It is also relevant for educators teaching these concepts and anyone interested in the foundational aspects of mathematical structures.

  • #31
PeroK said:
It's just the triangle inequality again:

##d(0, a) \le d(0, z) + d(z, a)##
##d(0, a) \le d(0, z) + d(z, a)## ⇔ ## d(0, z) ≥ d(0, a)-d(z, a) ##
 
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  • #32
lep11 said:
##d(0, a) \le d(0, z) + d(z, a)## ⇔ ## d(0, z) ≥ d(0, a)-d(z, a) ##

That's true,but what about using the inequalities you know for ##d(0, a)## and ##d(z, a)##?
 
  • #33
## d(0, z) ≥ d(0, a)-d(z,a) >d(0,a)>1 ##
 
  • #34
lep11 said:
## d(0, z) ≥ d(0, a)-d(z,a) > d(0,a)>1 ##

That middle equality cannot be correct. ##d(0, a)-d(z,a) \le d(0,a)## surely?
 
  • #35
PeroK said:
That middle equality cannot be correct. ##d(0, a)-d(z,a) \le d(0,a)## surely?
What's wrong?
 
  • #36
lep11 said:
What's wrong?

Come on! If you take a positive number away what you have gets smaller.
 
  • #37
PeroK said:
Come on! If you take a positive number away what you have gets smaller.
Okay, true.

I might just give up. I am not smart enough to study crap like this.
 
  • #38
lep11 said:
Okay, true.

I might just give up. I am not smart enough to study crap like this.
I took a break and tried again.

##d(0,z)≥d(0,x)-d(x,z)≥r+1-d(x,z)>r+1-r=1##
 
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  • #39
Thank You PeroK!
 

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