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lim inf (x_n) = inf {x: infinitely many x_n are < x }

lim sup (x_n) = sup {x: infinitely many x_n are > x }.

Now if both are equal to p, then consider a neighbourhood (p-e, p+e) of p (for some e>0).

As y:= p+ e/2 > p we cannot have that infinitely many x_n are > y (otherwise lim sup (x_n) >= y > p)

So at most finitely many x_n are > y, and almost all x_n are <= y < p+e.

Similarly with y' = y - e/2 and liminf: almost all x_n are >= y' > p-e.

So there is some N such that n >= N implies that x_n is in (p-e, p+e).

As e>0 was arbitrary x_n -- > x.

The reverse is similar, suppose x_n --> x.

If p > x, then there is some n such that x_n < p for all n >= N.

So p is NOT in the set B := {x: infinitely many x_n > x } so

B subset {t: t <= x} and so lim sup x_n = sup B <= x.

If p < x, then for all but finitely x_n, x_n >= p.

So p is not in A:= {t: infinitely many x_n < t} and so

A subset {t: t >= x } and so lim inf x_n = inf A >= x.

x <= liminf x_n <= limsup x_n <= x implies equality.

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# Homework Help: I cant visualize/understand this convergence proof

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