#Topology Question
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You are showing that the boundary of Q is R. That means, by definition, every r in R is a boundary point in Q.
Choose any point r in R and choose any e>0.
We must show that open ball of radius e and centered at r will contain both a rational and an irrational number in it. Do you know why?
Probably because on the coordinate axis, the left and right sides of a rational number are irrational numbers
Unfortunately, there isn't much sense in what you're saying, at least in the view of analysis and topology.
How would you proof that
We must show that every $r\in\mathbb{R}$ is a boundary point of $\mathbb{Q}$.
bonimy
So let's choose any $r\in\mathbb{R}$ and prove that it is a boundary point.
bonimy