Question #159250

Show that every open interval is an open set


1
Expert's answer
2021-02-04T07:40:49-0500

Let (i,j)(i,j) be an open interval. Let a(i,j)a \in (i, j) . We show that, we can find ϵ>0\epsilon > 0 \ni N(a,ϵ)(i,j)N(a, \epsilon) \sub (i,j)

Since a(i,j)a \in (i,j) we have that i<a<ji<a<j this implies that ai>0a-i > 0 and ja>0j-a > 0

Set ϵ=min(ai,ja)\epsilon = min (a-i, j-a)

Suppose further that bN(a,ϵ)b \in N(a, \epsilon) this implies that ba<ϵ|b-a| < \epsilon

ϵ<ba<ϵaϵ<b<a+ϵ- \epsilon < b - a< \epsilon \\ a - \epsilon < b < a + \epsilon

And since ϵ\epsilon is as defined we have that aϵia+ϵja - \epsilon \geq i \\ a + \epsilon \leq j

This implies that b(i,j)b \in (i, j)

Hence N(a,ϵ)(i,j)N(a, \epsilon ) \sub (i,j)


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