Given a sequence xn a Show that lim xn if and only if limin

Given a sequence {xn}. a) Show that lim xn = if and only if liminf xn = . b) Then show that lim xn = if and only if lim sup xn = . c) If {xn} is monotone increasing, show that either lim xn exists and is finite or lim xn = .

Exercise 2.3.14: Given a sequence tx a) Show that lim xn oo if and only if liminf x b) Then show that lim x oo if and only if lim sup x co. c) If txn is monotone increasing, show that either lim xin exists and is finite or lim xn oo.

Solution

Proof. If x = limxn, then the limit of every subsequence is also x so the “only if” direction is easy. Conversely, suppose that (xn)nN does not converge to x. Then there exists an open set U containing x such that xn / U for infinitely many n. Thus we can take a subsequence (xnk )kN for which xnk / U for all k N, and this subsequence clearly cannot have a further subsequence converging to x.

The proof of (b) is essentially identical. One direction is easy. Conversely, if lim xn = , there exists some R such that xn / (R, ) for infinitely many n. Then the subsequence of such terms notin in (R, ) clearly has no subsequence which converges to .

For (c), take any non-convergent sequence in a compact metric space X. Then any subsequence has a further convergent subsequence since X is compact and therefore sequentially compact, but the original sequence is non-convergent!