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HEINE iff CAUCHY
So I uploeded a picture of a theorem and it's proof, I understood the first part, however the second part I didn't understand why he just chose 1/n and wrote
xna<1/n, I mean from what I understand he did it so he could use the sandwich theorem and show that 0<xna<1/n , 1/n converges to 0 , zero converges to 0 and there fore xna converges to zero and we get xn>a, but because we assumed the that it is not continuous at a we got the sequence {f(xn)} doesn't converge to f(a).
But was it necessary to just assume xna<1/n ?
Firstly, if we just assume that for every xn that converges to a (in the range of the domain of the function), and we have the contrapositive of the continuity and therefore we get that f(xn) doesn't converge to f(a) and that's enough to prove without assuming xna<1/n... Isn't it? Or am I missing something here?
Secondly, I think assuming that xna<1/n is problematic because in this case we are talking about a specific situation that is when
xna<1/n . When we prove don't we have to talk about all sequences that converges to a, In other words, we need to prove for every sequence in general and not a specific situation when the sequence is smaller than 1/n. Because we are talking about the sequences only that are smaller than 1/n.
Appreciate if someone can give me feedback, I am clearly missing something here...
Thank you in advance.

Re: HEINE iff CAUCHY
Frankly, I don't think that the proof can be improved upon. One must fully understands what it means for $\displaystyle f$ not to be continuous at $\displaystyle a$.
http://www.texify.com/img/%5CLARGE%5...5Cright%5D.gif
Please note there is one $\displaystyle \varepsilon > 0$. But corresponding to any $\displaystyle \delta > 0$ there is at least one http://www.texify.com/img/%5CLARGE%5...5Cright%5D.gif
That is the reason that we can be assured that $\displaystyle x_n$ exists for each $\displaystyle \frac{1}{n}$.
EDIT: it should be http://www.texify.com/img/%5CLARGE%5...%5B/tex%5D.gif