.ANSWER.
1)An open interval is a set (a,b)={x∈R:a<x<b} .Let c∈(a,b) and ε=min{2b−c,2c−a} .ε<2b−c and −ε>2−c+a . Therefore
a=2a+a<2c+a=22c+a−c=c+2a−c <c−ε <c+ε<c+2b−c=2b+c<22b=b .
So, for all c∈(a,b) exist ε>0 , such that (c−ε ,c+ε)⊂(a,b) .By the definition of an open set , any open interval is an open set.
2)Similarly, the sets (−∞,a)={x∈R:−∞<x<a} and (b,+∞)=={x∈R:b<x<+∞} are open sets . Because , for all c∈ (−∞,a) exist ε=2−c+a,(c+ε=2c+a<2a+a=a) , such that (c−ε ,c+ε)⊂(−∞,a) .
If d∈(b,+∞)ε=2d−b(d−ε=2d+b>b) , then (d−ε,d+ε)⊂(b,+∞) .
3) A closed interval is a set [a,b]={x∈R:a≤x ≤b} =R╲((−∞,a)∪(b,+∞))
That is, the closed interval is the complement of the set I=(−∞,a)∪(b,+∞) .From 2) it follows that the set I is open. Therefore, by the definition of the closed set, the set
[a,b] is closed.
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