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13.4.1 If x ∈ E then let V = B(x, 12 ) = B(x, 12 ) ∩ E and W = E \ B(x, 12 ) = (X \ B(x, 12 )) ∩ E. Then V, W ⊆ E are open sets in the metric space (E, dE×E ) by Proposition 12.2.15. By the definition of the discrete metric we see that V = {x}, W = E \ {x}, so V ∪ W = E and V ∩ W = ∅. Also, as we assume E has at least two points, W is nonempty. Thus E is not connected.
13.4.2 By Theorem 13.4.6, we have that the continuous image of a connected set is connected. If f : X → Y is continuous and (X, d) is connected, then f (X) must be a connected subset of (Y, ddisc), which by the previous exercise means that f (X) can not have at least two points. Therefore f (X) contains only one point, so f is constant. Conversely suppose f is constant, i.e. f (x) = L for all x ∈ X. Then clearly whenever xn → x we have f (xn) → L = f (x). Therefore f is continuous.
13.4.6 We may reformulate connectedness in the following way: A set E is connected iff whenever E = U ∪V for disjoint relatively open sets U, V ⊆ E, either we have E = U or E = V. (This is just another way of saying that one of the sets must be empty.)
Let E = ∪α∈I Eα. We wish to show that E is connected. Suppose we write E = U ∪ V with U, V disjoint, relatively open subsets of E. By taking intersections with each of the Eα, we may then write each Eα as a disjoint union Uα ∪ Vα, where Uα = U ∩ Eα and Vα = V ∩ Eα are relatively open in Eα by Proposition 12.3.4. Since each Eα is a connected set, we must therefore have that either Eα ⊆ Uα ⊆ U or Eα ⊆ Vα ⊆ V.
Now we use the hypothesis that ∩α∈I Eα is nonempty. Because of this, we can choose some point x which lies in all the Eα. This x must be an element of U or of V. Without loss of generality, say it lies in U (if this is not the case we can switch the names of U and V to make it so). Then since each Eα is contained in either U or V , and each Eα contains the point x which is an element of U , we must have that all Eα are contained in U. Therefore E = U.
This proves that E is connected.
13.4.7 Suppose E is a path-connected subset of X, i.e. for any two points x, y ∈ E there exists a continuous map γ : [0, 1] → E such that γ(0) = x and γ(1) = y. Suppose E = U ∪ V , where U, V ⊆ E are disjoint, relatively open subsets of E. If U, V are nonempty we may choose x ∈ U and y ∈ V , and select a γ as above. But then since γ is continuous, we have that γ−^1 (U ), γ−^1 (V ) are disjoint open sets in [0,1] whose union is all of [0,1]. Since x = γ(0) and y = γ(1), we also have 0 ∈ γ−^1 (U ) and 1 ∈ γ−^1 (V ), so they are each nonempty sets. But then we have written [0,1] as a disjoint union of nonempty, open sets, which is impossible because we know that [0,1] is connected. Therefore E has no such decomposition, which means that E is connected.
14.1.1 We make the additional assumption that x 0 is an adherent point of the set E \ {x 0 }, so that lim x→x 0 ;x∈(E{x 0 })
f (x) does not fail to exist because of a technicality.
By definition, we have
lim x→x 0 ;x∈E f (x) = L ⇔ ∀ε > 0 ∃δ > 0 : d(x, x 0 ) < δ, x ∈ E ⇒ d(f (x), L) < ε
⇔ ∀ε > 0 ∃δ > 0 : x ∈ B(x 0 , δ) ∩ E ⇒ f (x) ∈ B(L, ε) (∗)
and
lim x→x 0 ;x∈(E{x 0 })
f (x) = L ⇔ ∀ε > 0 ∃δ > 0 : d(x, x 0 ) < δ, x ∈ (E{x 0 }) ⇒ d(f (x), L) < ε.
⇔ ∀ε > 0 ∃δ > 0 : x ∈ B(x 0 , δ) ∩ (E{x 0 }) ⇒ f (x) ∈ B(L, ε) (∗∗)
Since B(x 0 , δ) ∩ (E{x 0 }) ⊆ B(x 0 , δ) ∩ E if (∗) holds, then so does (∗∗). Therefore if the first limit is equal to L, then so is the second.
If the second limit is equal to f (x 0 ), we claim the first exists and is also equal to f (x 0 ). This is because given ε > 0, we can choose δ > 0 such that d(f (x), f (x 0 )) < ε for all x ∈ B(x 0 , δ) ∩ (E{x 0 }). Since d(f (x 0 ), f (x 0 )) = 0, we also have that d(f (x), f (x 0 )) < ε for all x ∈ B(x 0 , δ) ∩ E. This implies that the first limit is equal to f (x 0 ).
If the first limit exists, we claim it must be equal to f (x 0 ). This is because x 0 is contained in B(x 0 , δ) ∩ E for any δ > 0, so if the condition (∗) is to hold, we must have that for all ε > 0, d(f (x 0 ), L) < ε, which implies that d(f (x 0 ), L) = 0, so f (x 0 ) = L.
14.1.5 Let (xn)∞ n=1 be any sequence contained in E which converges to x 0. Since limx→x 0 ; x∈E f (x) = y 0 , we have by Proposition 14.1.5 that f (xn) converges to y 0. Clearly (f (xn))∞ n=1 is a sequence contained in f (E), so since limy→y 0 ; y∈f (E) g(y) = z 0 , we have that g(f (xn)) converges to z 0. Altogether, we have that for any sequence xn → x 0 contained in E, we have g(f (xn)) → z 0. Therefore limx→x 0 x∈E f (x) = z 0.
14.2.2 (a) If f (n)^ converges uniformly to f , then for all ε > 0 there exists N = N (ε) ∈ N such that for all n ≥ N and for all x ∈ X,we have d(f (n)(x), f (x)) < ε.
But clearly then, given x ∈ X and ε > 0, by choosing N = N (ε, x) = N (ε) in the same way as above, we have that for all n ≥ N , d(f (n)(x), f (x)) < ε, since the same is true for all points in X. Therefore f (n)^ converges to f pointwise.
(b) For any x ∈ (− 1 , 1) we have that |f (n)(x)| = |xn| = |x|n^ → 0 since |x| < 1, so the sequence f (n)(x) = xn^ converges to 0 pointwise.
If the sequence were to converge uniformly to any function it would have to be the 0 function, since uniform convergence implies pointwise convergence (by part (a)), and we already have pointwise convergence to the 0 function.
However, the sequence does not converge uniformly to 0. Indeed, for any N , choosing any n ≥ N , consider the sequence xk = 1 − (^) k^1. We have limk→∞(xk)n^ = (limk→∞ xk)n^ = 1n^ = 1 by the algebraic properties of limits. It follows from the definition of limits there exists an xk such that (xk)n^ > 12. This proves that d(f (n)(x), f (x)) cannot be uniformly less than 12 for any n, which rules out uniform convergence.
(c) For any given x ∈ (− 1 , 1), and any N ∈ N, we have that
1 + x + x^2 + · · · + xN^ = 1 − xN^ + 1 − x
