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Midterm solutions Robert Young November 2, 2015 1. Since `∞ is a metric space, B is compact if and only if it is sequentially compact. Let b1 = (1, 0, 0, . . . ) b2 = (0, 1, 0, . . . ) ... so that bi is the sequence with a 1 in the ith place and a 0 in every other place. Then d(bi , bj ), so the sequence (bi )∞ i=1 has no Cauchy subsequence and therefore has no convergent subsequence. ∞ On the other hand, if (ci )∞ i=1 is a sequence in C such that ci = (ci,j )j=1 , then by the compactness of the interval [−1, 1], there is a subsequence whose rst terms converge. Likewise, there is a subsequence of this whose second terms converge and so on. By a diagonal argument, we obtain a subsequence cni such whose nth terms converge for every n. We claim that this sequence converges in `∞ . 0 Denote the j th term of cni by c0i,j . Let L = (Li )∞ i=1 be the sequence of limits Lj = limi→∞ ci,j . Suppose that > 0. There is an N such that if i > N , then |c0i,j − Lj | < for all j < −1 . Further, since c0i,j , Lj ∈ [−1/j, 1/j], we have |c0i,j − Lj | < 2 for all j ≥ −1 . It follows that d(cni , L) < 2 for all i > N and thus that limi→∞ cni = L as desired. 2. If ai , i ∈ N is a sequence, then ai →` L if and only if ai converges to L in the standard topology and there are only nitely many i such that ai < L. If ai →` L, then for any neighborhood U of L, there is an N such that ai ∈ U for all i > N . Then, for any > 0, there is an N such that ai ∈ [L, L + ) ⊂ (L − , L + ) for all i > N . In particular, ai converges to L in the standard topology, and there are at most N terms in the sequence such that ai < L. Conversely, suppose that ai converges to L in the standard topology and that if i > N0 , then ai ≥ L. Let U = [c, d) be a basis element containing L. Then L < d, so there is an N such that ai < d for all i > N . If i > max{N0 , N }, then ai ∈ U . Thus, ai converges to L in R` . Suppose that A ⊂ R` is a set containing more than one point; i.e., c, d ∈ A, c < d. The sets [ [d, ∞) = [d, e) e>d and (−∞, d) = [ e<d 1 [e, d) are open in R` , so A ∩ [d, ∞) and A ∩ (−∞, d) are open in A. These are nonempty disjoint open sets that partition A, so A is disconnected. Therefore, the only connected subsets of R` are single points, so R` is totally disconnected. 3. Let X0 = X \ {p}. Then X0 is path-connected any two points can be connected by a U shaped path. If X = A ∪ B is a separation of X , then X0 ⊂ A or X0 ⊂ B ; suppose X0 ⊂ A. Since B is nonempty and disjoint from A, we must have B = {p} but any neighborhood of p intersects X0 , so B is not open. This is a contradiction, so X is connected. On the other hand, it is not path-connected. Suppose that γ : [0, 1] → X , γ(t) = (x(t), y(t)) is a continuous map such that γ(0) = p, γ(1) = (1, 0). Since γ is continuous, γ −1 (p) is a closed set, and since it is contained in [0, 1], it has a maximal element t0 . Clearly, t0 < 1. Let U = X0 \ L. This is an open set consisting of p and innitely many vertical lines. Since γ is continuous, there is an such that ([t0 , t0 + ]) ⊂ U . Then p = γ(t0 ) and p0 = γ(t0 + ) lie in the same connected component of U . But the connected component of p in U is just the point p, so p0 = p. This contradicts the maximality of t0 , so X is not path-connected. 4. Dene ∼ so that for all x, y ∈ X , x ∼ y if and only if $f(x)=f(y). Since f is continuous and constant on each equivalence class of ∼, it induces a continuous map g : X/ ∼→ Y given by g([x]) = f (x). Then g is injective, because if g([x]) = g([y]), then x ∼ y , and surjective, because f is surjective. Furthermore, since X/ ∼ is the continuous image of a compact set, X/ ∼ is compact. Therefore, g is a continuous bijection from a compact space to a Hausdor space. Any such map is a homeomorphism. 2