52 CHAPTER 2. BASIC TOPOLOGY AND ALGEBRA

Proof: Begin with (1). Let ε > 0 be given. Let xn→ x. Then by assumption, f(xn)→f(x) and g(xn)→ g(x) whenever xn→ x with xn ∈ D(f)∩D(g). Thus

|(af+bg)(xn)− (af+bg)(x)| → 0.

Now begin on (2). This also follows from properties of convergence of sequencesof real numbers from beginning calculus. For example, letting xn → x as above whereg(x) ̸= 0, |g(xn)−g(x)| < |g(x)|/2 for all n large enough and so, for all large enough n,from the triangle inequality, 3|g(x)|

2 ≥ |g(xn)| ≥ 12 |g(x)|∣∣∣∣ f (xn)

g(xn)− f (x)

g(x)

∣∣∣∣ =

∣∣∣∣ f (xn)g(x)− f (x)g(xn)

g(x)g(xn)

∣∣∣∣≤ 2

1

|g(x)|2| f (xn)g(x)− f (x)g(xn)|

≤ 21

|g(x)|2

(| f (xn)g(x)− f (x)g(x)|+ | f (x)g(x)− f (x)g(xn)|

)= 2

1

|g(x)|2

(|g(x)| | f (xn)− f (x)|+ | f (x)| |g(x)−g(xn)|

)which converges to 0 by assumption.

Now begin on (3). In terms of sequences, if xn→ x, then f(xn)→ f(x) and so

limn→∞

g(f(xn)) = g(f(x)) .

Thus g◦ f is continuous at x.Part (4) says: If f = ( f1, · · · , fq) : D(f)→Rq, then f is continuous if and only if each fk

is a continuous real valued function at x. Then letting xn→ x

max(| fi (xn)− fi (x)| , i = 1, ...,q)≤ |f(xn)− f(x)| ≡

(q

∑i=1| fi (xn)− fi (x)|2

)1/2

≤ √qmax(| fi (xn)− fi (x)| , i = 1, ...,q)

Thus f(xn)→ f(x) if and only if each fk (xn)→ f (x) and this shows (4).To verify part (5), the triangle inequality implies ||xn|− |x|| ≤ |xn−x| so if xn → x,

then |xn| → |x|. ■

2.5.2 Limits of FunctionsI will feature limits of functions which have values in some Rp. First of all, you can onlyconsider limits at limit points of the domain as explained below. It isn’t any harder toformulate this in terms of metric spaces, so this is what I will do. You can let the metricspace be Rp if you like.

Definition 2.5.23 Let f : D( f )⊆X→Y where (X ,d) and (Y,ρ) are metric spaces.For x a limit point of D( f ) , meaning that B(x,r) contains points of D( f ) other than x foreach r > 0, limy→x f (y) = z ∈ Y means the following.

For every ε > 0, there exists δ > 0 such that if 0 < d (x,y) < δ and y ∈ D( f ) , thenρ ( f (y) ,z)< ε .

52 CHAPTER 2. BASIC TOPOLOGY AND ALGEBRAProof: Begin with (1). Let € > 0 be given. Let x, — x. Then by assumption, f (x,) >f (x) and g(x,) > g(x) whenever x, — x with x, € D(f) 1 D(g). Thus|(af + bg) (Xn) — (af+ bg) (x)| > 0.Now begin on (2). This also follows from properties of convergence of sequencesof real numbers from beginning calculus. For example, letting x, — x as above whereg(x) £0, |g (Xn) — g(x)| < |g (x)| /2 for all n large enough and so, for all large enough n,from the triangle inequality, leo > |g (xn)| => 5 |g (x)|Fm) f)) _ slasdete)— fa)g(x) &(X) 8 (Xn)1Fp nde) —F 9) 8(a)1 (|) g(x) — F(x) g(x)<7w Gtx (x) - Fecal)Ig (x)| If (Xn) — f(x)marae ( + |F(x)| [8 () — 8 &n)| )which converges to 0 by assumption.Now begin on (3). In terms of sequences, if x,, — x, then f (x,) — f(x) and solim g(f(Xn)) = (f(x).n—s00Thus gof is continuous at x.Part (4) says: If f= (f1,--- , fg) : D(£) + RY, then f is continuous if and only if each fyis a continuous real valued function at x. Then letting x, — x1/2max (| fi (Xn) — fi(X)|,@=1,....9) S [£(%n) -f(x)| = wi=(¥ |fi (Xn) — fi (X)| *)< Jqmax (| fi (Xn) — fi (x)|,i= 1...)Thus f (x,) — f(x) if and only if each f, (x,) — f (x) and this shows (4).To verify part (5), the triangle inequality implies ||x,| — [x|| < |x, —x| so if x, > x,then |x,,| — |x|. Hi2.5.2 Limits of FunctionsI will feature limits of functions which have values in some R?. First of all, you can onlyconsider limits at limit points of the domain as explained below. It isn’t any harder toformulate this in terms of metric spaces, so this is what I will do. You can let the metricspace be R? if you like.Definition 2.5.23 Ler: D(f) CX + Y where (X,d) and (Y, Pp) are metric spaces.For x a limit point of D(f) , meaning that B(x,r) contains points of D(f) other than x foreach r > 0, limy_,x f (y) =z € Y means the following.For every € > 0, there exists 6 > 0 such that if 0 < d(x,y) < 6 and y € D(f), thenp(f(y),2) <&