Problem 3) Remark: for this problem, if I write the notation lim x n, it should always be assumed that I mean lim n x n, and similarly if I write the notation lim x nk it should always be assumed that I mean lim nk x nk a) Let x,y be Cauchy sequences in some metric space. Define d(x, y) = lim n d (x n, y n ). Show that this function is well-defined. Let the metric space in which our Cauchy sequences x,y are defined be denoted F. First, note that a distance function is defined from F F to R +. So for any two points a, b F, d(a, b) R +. Now, in order to show that d(x, y) is well defined, we have to show that the sequence (d(x n, y n )) n N is Cauchy, so lim n d (x n, y n ) converges in R. Ok, so pick some fixed arbitrary ɛ > 0 R. We want to show that N such that n, m > N = d(x n, y n ) d(x m, y m ) < ɛ. Note that for any m, n N, d(x n, y n ) d(x n, x m ) + d(x m, y n ) via the triangle inequality, which is d(x n, x m ) + d(x m, y n ) via the triangle inquality again, d(x n, x m ) + d(x m, y m ) + d(y n, y m ), by the triangle inequality used twice in 1 step; first to separate out the absolute values, and then to separate out the d s. Note that we can use the triangle inequality to split the d s and the absolute values because the d s represent a distance in F and the absolute values represent the distance of the sequence (d(a, b)) R Well, x is a Cauchy sequence, so by definition for any fixed value ɛ/2, N 1 such that n, m > N 1 = x n x m < ɛ/2, and similarly, since y is a Cauchy sequence, for any fixed value ɛ/2, N 2 such that n, m > N 2 = y n y m < ɛ/2. Letting N = max {N 1, N 2 }, we see that for n, m > N d(x n, y n ) d(x n, x m ) + d(x m, y m ) + d(y n, y m ) < ɛ/2+ d(x m, y m ) +ɛ/2 = d(x m, y m ) +ɛ. But by subtracting, this implies that d(x n, y n ) d(x m, y m ) < ɛ. But since distances in R are not negative, d(x n, y n ) = d(x n, y n ) and d(x m, y m ) = d(x m, y m ), so d(x n, y n ) d(x m, y m ) = d(x n, y n ) d(x m, y m ) < ɛ = d(x n, y n ) d(x m, y m ) < ɛ. This proves the claim that d(x, y) is a Cauchy sequence. So d(x,y) converges and is always well defined. b) Prove the antisymmetry axiom for our construction of the real numbers using Cauchy sequences. 1
Let us take two arbitrary Cauchy sequences x and y such that x y and x y. If x y then by the order defined in Harrison s lecture notes on Cauchy sequences, N 1 N such that for n > N 1 = x n y n. If also x y, then again by our definition of order on Cauchy sequences, N 2 N such that for n > N 2 = x n y n. So take N = max {N 1, N 2 }. Then for n > N, x n y n and x n y n, which implies x n = y n by antisymmetry in Q. So then for n > N, d(x n, y n ) = 0. It follows that lim n d(x n, y n ) = 0, since the sequence d(x n, y n ) becomes a sequence of 0 s after a certain cutoff N. This shows that x y.. c) Claim: Let x n be a Cauchy sequence with a subsequence x nk that converges to x. Then x n converges to x. Assume x nk converges to x. In a metric space, this means that lim nk d(x nk, x) = 0. We want to show that the limit of x n = x, which is the same (in a metric space) as saying that lim n d(x n, x) = 0. Well, we know that d(x n, x) d(x n, x nk )+d(x nk, x) by the triangle inequality. Now, any term x nk of the sequence (x nk ) is part of the overall sequence (x n ), so let us denote any term of the sequence (x nk ) by x m. Since the sequence (x n ) is Cauchy, N 1 such that n, m > N 1 = d(x n, x m ) < ɛ/2 by the definition of Cauchy sequence. So in particular, for our choice of n k = m, n, n k > N 1 = d(x n, x nk ) < ɛ/2 Furthermore, the sequence (x nk ) converges to x, so by the definition of convergence, for any x m (x nk ), N 2 such that m > N 2 = d(x m, x) < ɛ/2. So if we pick N = max {N 1, N 2 }, n > N = d(x n, x) d(x n, x nk ) + d(x nk, x) < ɛ/2 + ɛ/2 = ɛ. So then by definition, x n converges to x. Note: We did not denote d(a,b) as a b for Cauchy sequences a and b, since for problem c, the distance function of the metric space was not specified. d) Claim: The sequence x n converges to x every subsequence of x n converges to x. 1) ( = direction) 2
Claim: If a sequence x n converges, then every subsequence x nk converges to the same limit. First, note that n k k, k N. The proof of this is by induction. First, n k, k N is a subset of N because it is an indexing set of a subsequence, so for our base case, letting k = 1: n 1 1, since 1 is the lower bound of N. Inductive step: Inductive Hypothesis: suppose n i i, for some fixed arbitrary choice of i N. Well, n i+1 > n i, since these n k, k N are index for the subsequence x nk, so n i+1 denotes the term of the subsequence after n i. n i i by the inductive hypothesis, so n i+1 N > i, meaning n i+1 i+1, because i+1 is the next term in N. Thus, by induction, we have shown that n k k, k N. So now, let x = lim n x n, and let ɛ > 0. Then N such that n > N = x n x < ɛ. Then, k > N = n k > N by our earlier inductive argument, = x nk x < ɛ. So then by the definition of convergence, lim n x nk = x. 2) ( = direction) of x n converges to x. Then the set of sub- Suppose every subsequence x nk sequential limits S = x. We want to show three things now. Subclaim 1: lim sup x n S Subclaim 2: lim inf x n S. Subclaim 3: If lim sup x n = lim inf x n = x, then lim x n = x. Since S consists only of x, subclaim 1 and 2 show that lim sup x n = x = lim inf x n, so by subclaim 3, lim x n = x. So we have proven the claim. Subclaim 1 and 2: We need to construct subsequences x nk and x nj such that limsupx n = lim n x nk and liminfx n = lim n x nj. First, suppose x n is unbounded above. Then = limsupx n. Then let n 1 = 1. Given n 1 <... < n k, let n k+1 > max { k + 1, x nk+1 }. I can do this because x n is unbounded above. Then this subsequence is also unbounded 3
above,(i.e. lim n x nk = ), so S, meaning limsupx n S. Similarly, if x n is unbounded below, then = liminfx n. Let n 1 = 1, and given n 1 >... > n k, let n k+1 < min { k, x nk+1 }. I can do this because x n is unbounded below. Then this subsequence is also unbounded below,(i.e. lim n x nk = ), so S, meaning liminfx n S. The remaining cases are that x n is either bounded above or below. Since these cases are similar, without loss of generality, let x n be bounded above, (i.e. limsupx n = x, x ). Let ɛ > 0. Then by the definition of limsup, there are infinite points x n in the open epsilon ball around x. Let p be such a point, so let d(x, x n1 ) = ɛ 1 < ɛ. Then take ɛ 2 such that 0 < ɛ 2 < ɛ 1. There must be a point x n2 in this ball, since for some N, x sup {x n : n > N} < ɛ for any epsilon. Keep doing this to get a subsequence of x n that converges to x. Then limsupx n S. So we have proven subclaim 1, and w.l.o.g subclaim 2 as well. Remark: In Ross, pg 78, we find out that limsupx n = sups and liminfx n = infs. Subclaim 3: lim sup x n = x =lim inf x n. We want to show that lim x n = x. Let ɛ > 0. lim sup x n = lim N sup {x n : n > N} by definition. So we know there N1 N such that: x sup {x n : n > N1} < ɛ. This follows directly from the definition of limit. But then, this means precisely that x n < x + ɛ, n > N1 by manipulation of the inequality. In the same way, using the definition of lim inf, we know that since lim inf x n = x n = lim N inf {x n : n > N} by definition, N2 N such that: x inf {x n : n > N2} < ɛ,meaning that x n > x ɛ, n > N2 by manipulation of the inequality. So if we take N = max {N1, N2}, we get that x ɛ < x n < x + ɛ, n > N = n > N, x n x < ɛ, which means lim x n = x by definition. e) Claim: Let x n and y n be two real sequences. Suppose lim sup (x n ) < or lim sup (y n ) >. Then lim sup (x n + y n ) lim sup (x n ) + lim sup (y n ) Case 1: Suppose lim sup (x n ) = and lim sup (y n ) =. 4
Then lim sup (x n ) + lim sup (y n ) = + = (by definition). On the other hand, lim sup (x n + y n ) = lim n sup {x n + y n : n N} =, since for n N, the suprema of the x n and the suprema of the y n are both individually approaching. So since, lim sup (x n + y n ) lim sup (x n ) + lim sup (y n ). Case 2: Suppose lim sup (x n ) = and lim sup (y n ) =. Then lim sup (x n ) + lim sup (y n ) = = (by definition). On the other hand, lim sup (x n + y n ) = lim n sup {x n + y n : n N} =, since for n N, the suprema of the x n and the suprema of the y n are both individually approaching. So since, lim sup (x n + y n ) lim sup (x n ) + lim sup (y n ). Case 3: Suppose lim sup (x n ) < and lim sup (y n ) >. By definition of the supremum of a set, N N, x k sup {x n : n N} for all k N, and similarly, N N, y k sup {y n : n N} for all k N. Then x k +y k sup {x n : n N}+sup {y n : n N}, for all k N, meaning sup {x n : n N}+sup {y n : n N} is an upper bound for the set {x k + y k : k N}. Then sup {x n : n N} + sup {y n : n N} sup {x k + y k : k N}. Since this is true for all N N, as long as we have that k N, I can take limits as N on both sides and still preserve the equality. So: lim N (sup {x n : n N}+sup {y n : n N}) lim N (sup {x k + y k : k N}). But this is precisely an inequality for the lim sup of the three sequences, (x n ), (y n ) and (x n + y n ). So by definition, this implies that: limsup(x k + y k ) limsup(x k ) + limsup(y k ) 5