No, that's not true. The open interval $(a, b)$ is defined as the set of real numbers $x$ such that $a < x < b$; the half-open interval $[a, b)$ is defined as the set of real numbers $x$ such that $a \le x < b$. Here, we may allow $b = \infty$. Thus $(a,\infty)$ consists of those $x$ which satisfy $a < x < \infty$; similarly for $[a,\infty)$. For all real numbers $a$, $a\in [a,\infty)$ but $a\notin (a,\infty)$.
#38
ognik
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After some more fiddling with trying to grasp this, (from what you have explained, but also from what my prof said - he will be setting the end of year exam), I came up with the following:
If I was asked to write $ p \lt x \lt \infty $ as an interval, by definition I would write $ x \epsilon (p, \infty) $
If I was then told that $ p \gt 1 $ is a fixed parameter and that we had previously found that p is included in the interval, then I would write $ x \epsilon [p, \infty), p > 1 $ - I hope this is correct and well reasoned?
Then, if I wanted to try and write this semi-closed interval as an algebraic inequality, I would write $ 1 \lt p \lt x \lt \infty $, with p 'fixed' as P > 1
Have I got it?
#39
Euge
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You're so close! If you remove the "$x\in$" then you would be absolutely correct. [emoji2]
#40
ognik
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Thanks, Euge - but why remove the $ x \epsilon $ ... and do you mean from both places I have used it? ? x is the variable that we are trying to find the range for?
#41
Euge
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The notation $x\in (a, b)$ is not interval notation, but it express membership; it's $(a, b)$ itself that's interval notation. That's what I meant.
#42
ognik
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Hi Euge, I thought you might like to know, I found that the radius of convergence with the Weierstrauss test MUST be a closed interval, that helped. But what finally put my mind at rest, was to check convergence using the integral test - that showed me very clearly where the parameter (s) comes from.
A sincere thanks for your patience, it all finally came together for me - joy!