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 Show that there is a finite open cover.
+H) Sequential compact:
+∀ (y_n) in f(E), ∃ (y_n_k)_k s.t. lim y_n_k = y ∈ f(E) as k -> +inf
+I) If f: X -> R is continuous and E ⊂ X is compact, there exist p, q ∈ E s.t. f(p) = sup(f(E)), f(q) = inf(f(E))
+. How do we know K = (0, 1] is closed in (0, +inf)?
+J) Heine-Borel Theorem applies when X = R^n
+H) Pre-image of compact set may not be compact.
+Ex: f(x) = 1/x, Image = [0, 1]
+**Lecture 14**
+. Show that sinx is uniformly continuous.
+. Why is [0, 2π) not compact? (Referring to an example showing that if X is not compact, we cannot make a conclusion that if f: X -> Y is continuous and f is a bijection, the inverse is also continuous)
+A) Uniformly continuous (f: X -> Y):
+∀ ε > 0, ∃ δ > 0 s.t. for all p, q ∈ X, d_X(p, q) < δ -> d_Y(f(p), f(q)) < ε
+- One delta value works for all p ∈ X (unlike regular continuity)
+B) sinx is uniformly continuous, but x^2 is not
+C) If f: X -> Y is continuous and X is compact, f is uniformly continuous.
+D) If f: X -> Y is continuous, S ⊂ X, then f|_S: S -> Y is continuous.
+E) If f: X -> Y is continuous, X is compact, and f is a bijection, then f^-1: Y -> X is continuous.
+F) If f: X -> Y is uniformly continuous and S ⊂ X with the induced metric, then f|_S: S -> Y is also uniformly continuous.
+G) **Connected** space:
+The only subset of X that is both open and closed are X and ∅.
+H) If f: X -> Y is continuous, X is connected, then f(X) is connected.
+I) If f: X -> Y is continuous, E ⊂ X is connected, then f(E) is connected.
+**Lecture 15**
+A) Connected subset cannot be written as A ∪ B, where (closure of A) ∩ B = ∅ and A ∩ (closure of B) = ∅.
+. If the subset can be written as a union of A and B (the situation described above), how do we know that A, B are both open and closed in the subset?
+A: (closure of A) ∩ S = (closure of A) ∩ (A ∪ B) = ((closure of A) ∩ A) ∪ ((closure of A) ∩ B) = A ∪ ∅ = A. -> A and B are closed.
+A and B are complements. Thus, A and B are open.
+B) E is connected iff for all x, y ∈ E, and x < y, [x, y] ⊂ E.
+. Why does being a x ∈ X being a limit point of E ⊂ X imply that all (B_ε(x) \ {x}) ∩ E ≠ ∅?
+C) If f: (a, b) -> R is a monotone increasing function, f has at most countably many discontinuities.
+. If f: [0, 1] -> R is continuous and f([0, 1]) ⊂ [0, 1], there exists x ∈ [0, 1] s.t. f(x) = x. Prove this statement.
+**Lecture 16**
+. Give an example of a function that is pointwise convergent, but not uniformly convergent.
+A) Pointwise convergence (f_n: X -> Y):
+For all x ∈ X, lim f_n(x) = f(x) as n -> +inf
+B) Pointwise limit of a function does not preserve integral.
+. Describe the difference among Pointwise convergence, d2 convergence, and d∞ convergence.
+. How is d∞-metric sense convergence related to uniform convergence?
+A: f_n -> f uniformly iff lim d∞(f_n, f) = 0 as n -> +inf
+**Lecture 17**
+A) Uniformly continuous:
+for all ε > 0, there exists N > 0 s.t. for all n > N and for all x ∈ X, we have |f_n(x) - f(x)| < ε.
+- N only depends on ε, not on x.
+B) Uniformly Cauchy iff Uniformly convergent.
+C) f_n: X -> R, 0 <= M_n ∈ R s.t. M_n >= sup|f_n(x)| where x ∈ X.
+If an infinite series of M_n from n = 1 to n = +inf is less than +inf, then the infinite series of f_n from n = 1 to n = +inf converges uniformly.
+i.e. If the absolute value of f_n(x) is bounded (let's say the bound is M_n), and the infinite sum of M_n converges, then the infinite sum of f_n converges uniformly.
+D) Uniform convergence preserves continuity.
+E) To show that a function f is continuous, it suffices to show that for all x ∈ X', we have lim f(t) = f(x) as t -> x.
+. Why can we state the above (E)?
+F) If K is a compact metric space, f_n: K -> R, f_n is continuous, f_n -> f, f is continuous, and f_n(x) >= f_(n+1)(x),
+f_n -> f uniformly.
+**Lecture 18**
+A) K ⊂ X is compact iff K is closed and bounded **if X = R^n**
+Counterexample when X ≠ R^n:
+X = (0, 1), K = (0, 1)
+. Why is K not compact in the above example?
+B) Open and closed are relative notion, but compactness is an absolute notion.
+. If K ⊂ X is compact and E ⊂ X is closed, why are we able to conclude that K ∩ E is compact?
+C) Continuity preserves compactness and connectedness, but not necessarily openedness and closedness.
+. Give examples of (C) in which continuity does not preserve openedness that is not f(x) = x^2 and the domain is (-1, 1).
+. Show f(x) = sin(1/x) is not uniformly continuous.
+. How does f_n -> f uniformly translates to lim sup|f_n(x) - f(x)| = 0 as n -> +inf and for x ∈ X?
+. Show that f_n(x) = x/n converges to 0 pointwise, but not uniformly.
+D) E ⊂ X is dense iff (closure of E) = X
+**Lecture 19**
+A) If f is differentiable at p ∈ [a, b], then f is continuous at p.
+B) If f(x) is differentiable at p, then there exists u(x) s.t. f(x) = f(p) + (x-p)f'(p) + (x-p)u(x) where lim u(x) = 0 as x -> p
+u(x) = (f(x) - f(p))/(x - p) - f'(p) when x ≠ p, u(x) = 0 when x = p
+C) p can be a local maximum, but f'(p) is non existent (cusp).
+D) Local maximum and minimum can occur at endpoints.
+E) f: [a, b] -> R is a continuous function. Assume f'(x) exists for all x ∈ (a, b). If f(a) = f(b), then there exists c ∈ (a, b), s.t. f'(c) = 0. (Rolle's Theorem)
+F) If f, g: [a, b] -> R are continuous and differentiable on (a, b), then there exist c ∈ (a, b) s.t. [f(b) - f(a)]g'(c) = [g(b) - g(a)]f'(c).
+- Special case:
+f(b) - f(a) = (b-a)f'(c), c ∈ (a, b)
+**Lecture 20**
+. "If c is a local maximum of a function, and if derivative exists at c, and c is an interior point, derivative vanishes." How does this statement hold true?
+A) f: R -> R, f is continuous, f'(x) exists for all x ∈ R. Assume there exists M > 0 s.t. |f'(x)| <= M for all x. Then, f is uniformly continuous.
+B) f: [a, b] -> R is a differentiable function. f'(a) < f'(b). Then, for any c ∈ R w/ f'(a) < c < f'(b), there exists a d ∈ (a, b) s.t. f'(d) = c
+C) lim (f(x)/g(x)) = C as x -> a if:
+f, g: (a, b) -> R are differentiable, g(x), g'(x) ≠ 0 over (a, b) AND
+lim (f'(x)/g'(x)) = C AND
+i) lim f(x) = 0 as x -> a, lim g(x) = 0 as x -> a OR
+ii) lim g(x) = +inf as x -> a
+**Lecture 21**
+A) Smooth functions:
+Derivatives exist to all order.
+B) Taylor Theorem:
+f: [a, b] -> R is a function s.t. f^(n-1) (x) exists and is continuous on [a, b]. f^(n) (x) exists on (a, b).
+Then, for any α, β ∈ [a, b], we have:
+f(β) = f(α) + f'(α)(β-α) + f''(α)/2!(β-α)^2 + ... + f^(n-1)(α)/(n-1)!(β-α)^(n-1) + R_n(α, β)
+R_n(α, β) = 0 if α = β, R_n(α, β) = f^n(r)/n!(β-α)^n if α ≠ β for some r ∈ (α, β)
+**Lecture 22**
+A) Power series:
+Series of the form ∑ C_n(x-x_0)^n from n = 0 to n = +inf
+B) Radius of Convergence R:
+R = sup {r >= 0, s.t. if |x-x_0| <= r, the series converges}
+C) If R = 1/a, where a = limsup|C_n|^(1/n) as n -> +inf:
+If |x-x_0| < R, the series converges.
+If |x-x_0| > R, the series diverges.
+. Prove the diverging case of (C).
+D) Real Analytic function f:
+f: (a, b) -> R is smooth, for all x_0 ∈ (a, b), f(x) = ∑ C_n(x-x_0)^n for x in some neighborhood of x_0.
+. Regarding the definition of real analytic function, what does it mean for "f(x) = ∑ C_n(x-x_0)^n for x in some neighborhood of x_0"?
+E) Length of a function:
+∫sqrt(1 + (f'(x))^2)dx
+F) U(P, f) = ∑ M_i*Δx_i from i = 1 to i = n
+L(P, f) = ∑ m_i*Δx_i from i = 1 to i = n
+M_i = sup{f(x)|x ∈ [x_(i-1), x_i]}
+m_i = inf{f(x)|x ∈ [x_(i-1), x_i]}
+Δx_i = x_i - x_(i-1)
+G) U(f) = inf U(P, f)
+L(f) = sup L(P, f)
+H) f is integrable if U(f) = L(f)
+I) Generalization of Riemann-Stieltjes integrable:
+Let α: [a, b] -> R be a monotone increasing function, define partition P = {a = x_0 <= x_1 <= ... <= x_n = b}, define Δα_i = α(x_i) - α(x_(i-1))
+The remaining definitions are similar as in parts (F) and (G), except Δx_i -> Δα_i, and U(P, α) = L(P, α) implies that f is Riemann-Stieltjes integrable w.r.t. α.
+**Lecture 23**
+A) If a partition Q is a refinement of partition P on [a, b], then L_P <= L_Q <= U_Q <= U_P.
+B) L(f, α) <= U(f, α)
+C) f is integrable w.r.t. α iff for all ε > 0, there exists P partition s.t. U_P - L_P < ε
+.
+General Note:
+A) Contradiction is very useful in proofs.
+When using contradiction, you can select an arbitrary element in a set and prove if it actually belongs to a set.

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