Problem Set #2 Part C Official Solutions

Inna Zakharevich

October 11, 2003

1.

(10) (i) We will first show that if f is continuous and {x_{n}} converges to x then {f(x_{n})} → ∞ as n → ∞. Since f is continuous at x we know that for every ϵ > 0 there exists a δ > 0 such that if d(x, x′) < δ then d(f(x), f(x′)) < ϵ. Note that since {x_{n}} → x then for δ there exists an integer N > 0 such that if n ≥ N then d(x_{n}, x) < δ. From this we see that if n ≥ N, d(f(x_{n}), f(x)) < ϵ, so {f(x_{n})} → f(x). Now we need to show that if every sequence such that {x_{n}} → x also has {f(x_{n})} → f(x) then f is continuous. Solution 1: Suppose that there is some x such that f is not continuous at x. That means that there exists ϵ > 0 such that for all δ > 0 there exists x′ such that d(x, x′) < δ but d(f(x), f(x′)) ≥ ϵ. Define a sequence {x_{n}} in the following way: setting δ = 1/n let x_{n} be such that d(x, x_{n}) < δ but d(f(x), f(x_{n})) ≥ ϵ. Then this sequence converges to x. However, it does not converge to f(x), since f(x_{n}) is at least ϵ from f(x) for all n. Contradiction. Thus f must be continuous. Solution 2: We know that f is continuous if and only if the preimage of any closed set is closed. Let F ⊂ Y be closed. We will show that f^{−1}(F) is closed. Consider any sequence {x_{n}} in f^{−1}(F). If we can show that x ∈ f^{−1}(F) we will be done, since a closed set is exactly one that contains all of its limit points. Consider {f(x_{n})} in F. F is closed, so we know that f(x) is in F, so x ∈ f^{−1}(F). Thus f^{−1}(F) is closed, so f is continuous.

(10) (ii) This function is continuous at every irrational and discontinuous at every rational. First we will show that it is discontinuous at every rational. Let {x_{n}} be a sequence of irrationals converging to any rational p/q. Then {f(x_{n})} is identically 0. Thus it converges to 0. However, f(p/q) = 1/q > 0, so f is discontinuous at every rational. Now consider any irrational x ∈ [0, 1]. Consider any sequence {x_{n}} converging to x. If it has only a finite number of rationals then we can simply consider the subsequence that consists only of irrationals; this will converge to the limit of the sequence. Then {f(x_{n})} will clearly converge to 0, which is f(x). Now suppose that the sequence has infinitely many rationals. We can split this sequence up into two: one with all irrationals, one with all rationals. Let {y_{n}} be the subsequence of rationals, and {z_{n}} be the subsequence of rationals. If {y_{n}} is finite we can disregard it. If it does not, we have already shown that {f(y_{n})} will converge to 0. Thus all we need to show is that {f(z_{n})} will converge to 0. Fix ϵ > 0. Let N > 1/ϵ, and let δ = min{|p/q−x| : q < N, p ≤ q}. Let M be such that if m > M then d(y_{m}, x) < δ. Then note that f(y_{m}) < 1/N < ϵ, so d(f(y_{m}), f(x)) < ϵ. Thus {f(y_{n})} converges to 0, so we see that f is continuous at every irrational x.

(10) 2. We will show that R^{2}\{x_{1}, …, x_{n}} is path-connected. Let r = 1/3 min{|x_{i} − x_{j}| : 1 ≤ i, j ≤ n}. Note that any two circles of radius r centered at two of the removed points do not intersect. Pick any two points in x, y ∈ R^{2}\{x_{1}, …, x_{n}}. Connect them with a straight line. If this line does not contain any point in {x_{1}, …, x_{n}} then we are done. Suppose that this line contains some point x_{j} ∈ {x_{1}, …, x_{n}}, draw C, the circle of radius r centered around x_{j}. Then the line will intersect C at two points. Replace the segment inside C with one of the two arcs connecting the two points of intersection. Then this path can clearly be described by a continuous function from [0, 1] into R^{2}\{x_{1}, …, x_{n}}. Thus any two points are connected, so R^{2}\{x_{1}, …, x_{n}} is connected.

(10) 4. Let f be a homeomorphism X → Y. Note that for any closed set Z ∈ X f is also a homeomorphism from X\Z to Y \f(Z).

First consider the circle and the interval in R. Suppose that there is some homeomorphism f between them. First note that since homeomorphisms preserve compactness, we know that the interval must be closed, since the circle is compact. Now consider a point x on the interior on the interval. Let Z = {x}. Then f will be a homeomorphism between two disjoint intervals in R and a circle with a point removed. However, note that the circle is still connected, and the interval is not. Thus f cannot be a homeomorphism.

Now consider the figure-8 and the circle. Suppose that there is a homeomorphism f between them. Let x be the point of intersection of the figure-8. Then the figure-8 minus x is disconnected, while the circle minus f(x) is still connected. However, f would be a homeomorphism between these two sets, which is a contradiction since homemorphisms preserve conectedness. Thus the circle is not homeomorphic to the figure-8.