← 4. Continuity

Exercises

Problems to check your understanding of this chapter. The solutions are folded away — try them yourself first.

Easy

1EasyCounterexample

True or false, with proof: the continuous image of an open interval is always an open set.

f(x)=x2,I=(−1,1),f(I)=[0,1)f(x) = x^2, \qquad I = (-1, 1), \qquad f(I) = [0, 1)
Verdict

The claim is false — the image of the open interval I=(−1,1)I = (-1, 1) under f(x)=x2f(x) = x^2 is [0,1)[0, 1), which is not an open set.

Show solution

Counterexample and verification

  1. False. f(x)=x2f(x) = x^2 on R\mathbb{R} is continuous and I=(−1,1)I = (-1, 1) is an open interval. We claim f(I)=[0,1)f(I) = [0, 1).

    If 0≤y<10 \le y < 1 then x=y∈[0,1)⊂Ix = \sqrt{y} \in [0, 1) \subset I and f(x)=yf(x) = y, so y∈f(I)y \in f(I). If y<0y < 0 it is unreachable since x2≥0x^2 \ge 0; if y≥1y \ge 1 it is unreachable since ∣x∣<1|x| < 1 implies x2<1x^2 < 1.

  2. [0,1)[0, 1) is not open in R\mathbb{R}: 0∈[0,1)0 \in [0, 1) but for every ε>0\varepsilon > 0, −ε/2∈(−ε,ε)-\varepsilon/2 \in (-\varepsilon, \varepsilon) while −ε/2∉[0,1)-\varepsilon/2 \notin [0, 1).

    So the continuous image [0,1)[0, 1) of the open interval is not open. Continuity does not preserve openness (it does preserve closedness via compactness: the continuous image of a compact set is compact).

Medium

2Medium

On R\mathbb{R} let fn(x)=x1+nx2f_n(x) = \frac{x}{1 + n x^2}.

Find the pointwise limit lim⁡n→∞fn(x)\lim_{n \to \infty} f_n(x) at each xx, prove the convergence is uniform on R\mathbb{R}, and compute the exact value of sup⁡x∈R∣fn(x)∣\sup_{x \in \mathbb{R}} |f_n(x)|.

fn(x)→0(n→∞) for every xf_n(x) \to 0 \quad (n \to \infty) \text{ for every } x
sup⁡R∣fn∣=12n→0\sup_{\mathbb{R}} |f_n| = \frac{1}{2\sqrt{n}} \to 0
Show solution

  1. First the pointwise limit. If x=0x = 0 then fn(0)=0f_n(0) = 0.

    If x≠0x \ne 0 then 1+nx2≥nx21 + nx^2 \ge nx^2, so ∣fn(x)∣≤∣x∣nx2=1n∣x∣→0|f_n(x)| \le \frac{|x|}{nx^2} = \frac{1}{n|x|} \to 0. Hence the pointwise limit is f≡0f \equiv 0.

  2. Uniform convergence is decided by the sup norm: fn→ff_n \to f uniformly iff sup⁡∣fn−f∣→0\sup |f_n - f| \to 0. Since fnf_n is odd, the supremum of ∣fn∣|f_n| is found on x≥0x \ge 0.

    Differentiating there, gn′(x)=1−nx2(1+nx2)2g_n'(x) = \frac{1 - nx^2}{(1 + nx^2)^2}, so the unique critical point is x=1/nx = 1/\sqrt{n} with value gn(1/n)=1/n2=12ng_n(1/\sqrt{n}) = \frac{1/\sqrt{n}}{2} = \frac{1}{2\sqrt{n}}. The boundary values (x=0x = 0 and x→∞x \to \infty) are 00, so this is the global maximum.

  3. By odd symmetry, sup⁡x∈R∣fn(x)∣=12n→0\sup_{x \in \mathbb{R}} |f_n(x)| = \frac{1}{2\sqrt{n}} \to 0. Hence fn→0f_n \to 0 uniformly on R\mathbb{R}.

    The bump's location 1/n1/\sqrt{n} drifts toward 00, but its height 1/(2n)1/(2\sqrt{n}) collapses too, which is exactly why the convergence is uniform.