By S. B. Kizlik, Michael D. Adam
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V) If lim x→a f (x) = 0, then lim x→a 1f (x) = 1 . lim x→a f (x) (vi) If a ∈ dom( f ), f (a) = 0 and f is continuous at a, then a. 5. 4. (iv) the coordinate functions Rn → R with x → x j , for 1 ≤ j ≤ n, are continuous. It then follows from part (iv) in the corollary above and by mathematical induction that monomial functions Rn → R, that is, functions of the form x → x1α1 · · · xnαn , with α j ∈ N0 for 1 ≤ j ≤ n, are continuous. In turn, this fact and part (ii) of the corollary imply that polynomial functions on Rn , which by definition are linear combinations of monomial functions on Rn , are continuous.
Assume K not bounded. Then we can find a sequence (xk )k∈N satisfying xk ∈ K and xk ≥ k, for k ∈ N. Obviously, in this case the extraction of a convergent subsequence is impossible, so that K cannot be compact. 12. Indeed, all subsequences of a convergent sequence converge to the same limit. ❏ It follows from the two preceding theorems that [ −1, 1 ] and R are not homeomorphic; indeed, the former set is compact while the latter is not. 3. (iii)); nevertheless, the inverse image of a compact set might fail to be compact.
Ii) Homogeneity: ν(λx) = |λ| ν(x). (iii) Triangle inequality: ν(x + y) ≤ ν(x) + ν(y). 10 (Equivalence of norms on Rn ). For every norm ν on Rn there exist numbers c1 > 0 and c2 > 0 such that c1 x ≤ ν(x) ≤ c2 x (x ∈ Rn ). In particular, ν is Lipschitz continuous, with c2 being a Lipschitz constant for ν. 28 Chapter 1. Continuity Proof. We begin by proving the second inequality. 6) ν(x) = ν xjej ≤ 1≤ j≤n |x j | ν(e j ) ≤ x (ν(e1 ), . . , ν(en )) =: c2 x . 1≤ j≤n Next we show that ν is Lipschitz continuous.