MHB Decreasing nonnegative sequence and nonincreasing functions

ozkan12
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Let $\{p_n\}$ be a nonnegative nonincreasing sequence and converges to $p \ge 0$. Let $f : [0,\infty)\to[0,\infty)$ be a nondecreasing function. So, since f is a nondecreasing function, $f(p_n)>f(p)>0$. How did this happen?
 
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Hi ozkan12,

Since $\{p_n\}$ is nonincreasing and converging to $p$, $p = \inf\{p_n : n\in \Bbb N\}$, so $p_n \ge p$ for all $n\in \Bbb N$. Consequently, as $f$ is nondecreasing, $f(p_n) \ge f(p)$. So we have $f(p_n) \ge f(p) \ge 0$.

Note that the inequalities above are not strict.
 
Euge said:
Hi ozkan12,

Since $\{p_n\}$ is nonincreasing and converging to $p$, $p = \inf\{p_n : n\in \Bbb N\}$, so $p_n \ge p$ for all $n\in \Bbb N$. Consequently, as $f$ is nondecreasing, $f(p_n) \ge f(p)$. So we have $f(p_n) \ge f(p) \ge 0$.

Note that the inequalities above are not strict.

ok thanks a lot :) but I will ask one question : so, pn+1<=pn-f(p)...how this happened ?
 
ozkan12 said:
ok thanks a lot :) but I will ask one question : so, pn+1<=pn-f(p)...how this happened ?

I'm sorry, I don't understand -- where did you get $p_{n+1} \le p_n - f(p)$? This need not be true.
 
Euge said:
I'm sorry, I don't understand -- where did you get $p_{n+1} \le p_n - f(p)$? This need not be true.

ok :) I see my false :) I repair that :)
 
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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