lep11
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##d(0, a) \le d(0, z) + d(z, a)## ⇔ ## d(0, z) ≥ d(0, a)-d(z, a) ##PeroK said:It's just the triangle inequality again:
##d(0, a) \le d(0, z) + d(z, a)##
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.
PREREQUISITESStudents 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.
##d(0, a) \le d(0, z) + d(z, a)## ⇔ ## d(0, z) ≥ d(0, a)-d(z, a) ##PeroK said:It's just the triangle inequality again:
##d(0, a) \le d(0, z) + d(z, a)##
lep11 said:##d(0, a) \le d(0, z) + d(z, a)## ⇔ ## d(0, z) ≥ d(0, a)-d(z, a) ##
lep11 said:## d(0, z) ≥ d(0, a)-d(z,a) > d(0,a)>1 ##
What's wrong?PeroK said:That middle equality cannot be correct. ##d(0, a)-d(z,a) \le d(0,a)## surely?
lep11 said:What's wrong?
Okay, true.PeroK said:Come on! If you take a positive number away what you have gets smaller.
I took a break and tried again.lep11 said:Okay, true.
I might just give up. I am not smart enough to study crap like this.