Cant understand this step in a bounded prove

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The discussion centers on proving that a function f(x) is bounded on the interval (-∞, a] given that the limit as x approaches -∞ exists and is finite. Specifically, if lim_{x->-∞} f(x) = m_0 and m_0 < a, the proof involves demonstrating that if f(x) is not bounded from above, then for every natural number n, there exists an x_n within the interval [m_0, a] such that m_0 ≤ x_n ≤ a. This leads to the conclusion that the supremum of f(x) on (-∞, a] is indeed the maximum value.

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transgalactic
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the question:
f(x) continues on [tex](-\infty,a][/tex]
and suppose that the border [tex]\lim_{x->-\infty}f(x)[/tex] exists and finite.
prove that f(x) is bounded on [tex](-\infty,a][/tex] and/or that exists
[tex]x_0\epsilon(-\infty,a]=\lim _{x->-\infty }f(x)[/tex]
so
[tex]\sup_{x\epsilon(-\infty,a]} f(x)[/tex]
in other words prove that f(x) gets its highest value on [tex](-\infty,a][/tex]
and the supremum is the maximum

the non understood part:

suppose
[tex]\lim_{x->-\infty}f(x)=m_0[/tex]
suppose [tex]m_0<a[/tex]
and we check on the interval of [tex][m_0,a][/tex] where [tex][m_0,a]\subseteq (-\infty,a][/tex]

they prove by a counter example that:
"suppose the function is not bounded from the top then [tex]\forall n\epsilon N[/tex] and
[tex]m_0\leq x_n\leq a[/tex]"

i can't understand it.if a function is bounded by some epsilon then we take N for which after this N (n>N) f(x)<epsilon
if its not bounded from the top then
f(x) is bigger then epsilon for the whole interval

this is not what they writee up there
what are they writing there??
 
Last edited:
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What's xn, and why are you looking at all of them? I'm not sure I understand what's going on in your post
 
how its supposed to be diverged?
 

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