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moxy
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Homework Statement
I am to prove that
[itex]\lim_{x\rightarrow t}{f(x)} = L \Longleftrightarrow \limsup{f(t)} = \liminf{f(t)} = L[/itex]
where I is an interval in ℝ, and when [itex]f:I → ℝ[/itex], then [itex]\forall t \in I[/itex] we let [itex]N_t[/itex] denote the set of all neighborhoods of t relative to I, and we define:
[itex]\limsup{f(t)} := \inf_{U \in N_t} \sup{\{f(s) | s \in U\}}[/itex]
[itex]\liminf{f(t)} := \sup_{U \in N_t} \inf{\{f(s) | s \in U\}}[/itex]
The Attempt at a Solution
I'm working on [itex]\lim_{x\rightarrow t}{f(x)} = L \Longleftarrow \limsup{f(t)} = \liminf{f(t)} = L[/itex] at the moment.
So, I've assumed [itex]\limsup{f(t)} = \liminf{f(t)} = L[/itex].
That is,
[itex]\inf_{U \in N_t} \sup{\{f(s) | s \in U\}} := \inf_{U \in N_t} \sup{\{f(U)\}} = L[/itex] and
[itex]\sup_{U \in N_t} \inf{\{f(s) | s \in U\}} := \sup_{U \in N_t} \inf{\{f(U)\}} = L[/itex]
I know that sup(f(U)) ≥ L for any U in Nt, so there exists some U1 in Nt such that sup(f(U1)) - ε > L for any ε > 0.
==> sup(f(U1)) > L + ε for all ε > 0
Also, inf(f(U)) ≤ L for any U in Nt, so clearly inf(f(U1)) ≤ L < L + ε for any ε > 0.
sup(f(U1)) > L + ε
inf(f(U1)) < L + εI don't think that last step is actually getting me anywhere...I'm completely stuck. Any suggestions would be greatly appreciated. My prof recently introduced liminf, limsup, and neighborhoods, and the new terminology is really messing with my head.
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