MHB Density of Q in R .... Sohrab Theorem 2.1.38 ....

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The discussion revolves around understanding the assumption that x > 0 in the proof of Theorem 2.1.38 from Sohrab's "Basic Real Analysis." It is clarified that for a sufficiently large integer k, the expression x + k will always be positive. This allows for the identification of a rational number r between x + k and y + k, ensuring that r - k is a rational number between x and y. The explanation emphasizes the importance of this assumption in the context of the theorem's proof. Overall, the conversation highlights the logical reasoning behind the assumption in the theorem.
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I am reading Houshang H. Sohrab's book: "Basic Real Analysis" (Second Edition).

I am focused on Chapter 2: Sequences and Series of Real Numbers ... ...

I need help with an aspect of the proof of Theorem 2.1.38 ...

Theorem 2.1.38 reads as follows:https://www.physicsforums.com/attachments/7089In the above text by Sohrab, at the start of the proof, we read the following:

"We may assume that $$x \gt 0$$. (Why) ... ... "Can someone please explain to me why we can assume that $$x \gt 0$$?

Peter
 
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Peter said:
Can someone please explain to me why we can assume that $$x \gt 0$$?Peter
For a sufficiently large integer $k$, $x+k$ will be positive. If you can find a rational number $r$ between $x+k$ and $y+k$, then $r-k$ will be a rational number between $x$ and $y$.
 
Opalg said:
For a sufficiently large integer $k$, $x+k$ will be positive. If you can find a rational number $r$ between $x+k$ and $y+k$, then $r-k$ will be a rational number between $x$ and $y$.
Thanks Opalg ...

Easy when you see how it works! ... :) ...

... appreciate the help ...

Peter
 
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We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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