Kenneth Kuttler

101

Notice that the concept of continuity as described in the definition is a point property.That is to say it is a property which a function may or may not have at a single point. Hereis an example.

Example 6.0.3 Let

f (x) ={

x if x is rational0 if x is irrational .

This function is continuous at x = 0 and nowhere else.

If xn → 0, then | f (xn)| ≤ |xn| and |xn| → 0 so it follows that f (xn)→ 0 ≡ f (0) andso the function is continuous at 0. However, if x ̸= 0 and rational, you could consider asequence of irrational numbers converging to x, {xn} and f (xn) = 0→ 0 ̸= f (x). If xis irrational, you could pick a sequence of rational numbers {xn} converging to x and sof (xn) = xn→ x ̸= f (x). Here is another example.

Example 6.0.4 Show the function f (x) =−5x+10 is continuous at x = −3.

To do this, note first that f (−3) = 25 and it is desired to verify the conditions forcontinuity. Consider the following. |−5x+10− (25)|= 5 |x− (−3)| .

This allows one to find a suitable δ . If ε > 0 is given, let 0 < δ ≤ 15 ε. Then if 0 <

|x− (−3)| < δ , it follows from this inequality that |−5x+10− (25)| = 5 |x− (−3)| <5 1

5 ε = ε.Sometimes the determination of δ in the verification of continuity can be a little more

involved. Here is another example.

Example 6.0.5 Show the function f (x) =√

2x+12 is continuous at x = 5.

First note f (5) =√

22. Now consider:∣∣∣√2x+12−

√22∣∣∣= ∣∣∣ 2x+12−22√

2x+12+√

∣∣∣=

2√

2x+12+√

22|x−5| ≤ 1

√22 |x−5|

whenever |x−5|< 1 because for such x,√

2x+12 > 0. Now let ε > 0 be given. Choose δ

such that 0 < δ ≤min(

1, ε√

222

). Then if |x−5|< δ , all the inequalities above hold and

∣∣∣√2x+12−√

22∣∣∣≤ 2√

22|x−5|< 2√

22ε√

222

= ε.

Example 6.0.6 Show f (x) =−3x2 +7 is continuous at x = 7.

Suppose xn→ x. Then by the theorem on limits, Theorem 4.4.8, −3x2n +7→−3x2 +7

and so this function is continuous at x. In particular, it is continuous at 7.

Proposition 6.0.7 For x ∈ Fp, and S⊆ Fp,S ̸= /0, let

inf{∥x− s∥ : s ∈ S} ≡ dist(x,S)

Then|dist(x,S)−dist(y,S)| ≤ ∥x− y∥ (∗)

so dist : Fp→ R is continuous.

101Notice that the concept of continuity as described in the definition is a point property.That is to say it is a property which a function may or may not have at a single point. Hereis an example.Example 6.0.3 Letx if x is rational0 if x is irrational ~fis)={This function is continuous at x = 0 and nowhere else.If x, — 0, then |f (x,)| < |x| and |x,| > 0 so it follows that f (x,) + 0 = f (0) andso the function is continuous at 0. However, if x 4 0 and rational, you could consider asequence of irrational numbers converging to x, {x,} and f(x,) =0 + OF f(x). If xis irrational, you could pick a sequence of rational numbers {x,} converging to x and sof (Xn) =X 2 x # f (x). Here is another example.Example 6.0.4 Show the function f (x) = —5x+ 10 is continuous at x = —3.To do this, note first that f(—3) = 25 and it is desired to verify the conditions forcontinuity. Consider the following. |—5x + 10 — (25)| =5|x—(—3)|.This allows one to find a suitable 6. If € > 0 is given, let0< 6 < Zé. Then if 0 <|x —(—3)| < 6, it follows from this inequality that |—5x+10—(25)| = 5|x—(—3)| <StE=E.Sometimes the determination of 6 in the verification of continuity can be a little moreinvolved. Here is another example.Example 6.0.5 Show the function f (x) = /2x+ 12 is continuous at x =5.First note f (5) = 22. Now consider: | V2e+ 12— v2 = | ea2 1= ————_—— _ |x—-5|] < — V22|x—-5eel+ Vo [<p 77h 5whenever |x —5| < 1 because for such x, /2x+ 12 > 0. Now let € > 0 be given. Choose 6such that 0 < 6 < min (1. ev?) . Then if |x —5| < 6, all the inequalities above hold and22 2 2/22V2x+12— V2] < ——|x—5 <— ~¢| Syl V2.2Example 6.0.6 Show f (x) = —3x* +7 is continuous at x = 7.Suppose x, > x. Then by the theorem on limits, Theorem 4.4.8, —3x2 +7 —3x° +7and so this function is continuous at x. In particular, it is continuous at 7.Proposition 6.0.7 For x € F’, and S CF’,S #9, letinf {||x — || : s € S} = dist (x, S)Then|dist (x, 8) — dist (y,S)| < |x| (*)so dist : F? > R is continuous.