Question

Prove that X is totally bounded if every sequence of X has a convergent subsequence. Please directly prove it without using any theorem on totally boundedness.

Answer #1

Prove that every bounded sequence has a convergent
subsequence.

Suppose that every Cauchy sequence of X has a convergent
subsequence in X. Show that X is complete.

Show that sequence {sn} converges if it is monotone
and has a convergent subsequence.

Find an example of a sequence, {xn}, that does
not converge, but has a convergent subsequence.
Explain why {xn} (the divergent sequence) must have an
infinite number of convergent subsequences.

If (x_n) is a convergent sequence prove that (x_n) is bounded.
That is, show that there exists C>0 such that abs(x_n) is less
than or equal to C for all n in naturals

Prove: If x is a sequence of real numbers that converges to L,
then any subsequence of x converges to L.

Suppose (an) is an increasing sequence of real numbers. Show, if
(an) has a bounded subsequence, then (an) converges; and (an)
diverges to infinity if and only if (an) has an unbounded
subsequence.

(a) Prove that the sum of uniformly convergent sequences is also
a uniformly convergent sequence.
(b) Prove that if, in addition to part (a), the sequences are
bounded, then the product is also uniformly convergent.

Prove that if a sequence converges to a limit x then very
subsequence converges to x.

Prove Corollary 4.22: A set of real numbers E is closed and
bounded if and only if every infinite subset of E has a point of
accumulation that belongs to E.
Use Theorem 4.21: [Bolzano-Weierstrass Property] A set of real
numbers is closed and bounded if and only if every sequence of
points chosen from the set has a subsequence that converges to a
point that belongs to E.
Must use Theorem 4.21 to prove Corollary 4.22 and there should...

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