Analysis I: Lecture 7

In which we meet the definition of continuity for functions.

Definition of what it means for a complex-valued function to be continuous at a point and on a set.
Lemma 14: Let $E$ be a subset of $\mathbb{C}$ , and let $f$ and $g$ be functions from $E$ to $\mathbb{C}$ . Let $a$ be a point in $E$ .
1. If there is some $c$ such that $f(z) = c$ for all $z$ in $E$ , then $f$ is continuous on $E$ .
2. If $f$ and $g$ are continuous at $a$ , then so is the pointwise sum $f+g$ .
3. If $f$ and $g$ are continuous at $a$ , then so is the pointwise product $f \times g$ .
4. Suppose that $f(z) \neq 0$ for all $z$ in $E$ . If $f$ is continuous at $a$ , then so is $1/f$ .
Exercise: show that the function $f:E \to \mathbb{C}$ given by $f(z) = z$ is continuous on $E$ . Together with Lemma 14, this shows that polynomials are continuous.
Lemma 15: Let $E$ be a subset of $\mathbb{C}$ . Let $f$ be a function from $E$ to $\mathbb{C}$ that is continuous at some point $a$ in $E$ . Let $(z_n)_{n=1}^{\infty}$ be a sequence in $E$ such that $z_n \to a$ as $n \to \infty$ . Then $f(z_n) \to f(a)$ as $n \to \infty$ .

Understanding today’s lecture

Define the function $f: \mathbb{R} \to \mathbb{R}$ by setting $f(x) = x$ if $x$ is rational and $f(x) = 0$ if $x$ is irrational. At which points is $f$ continuous?

Pick some functions and check whether you can use the definition of continuity to show that they are/aren’t continuous. (We haven’t officially defined functions such as $\exp$ , $\log$ , $\sin$ and $\cos$ in this course, so you might want to steer clear of those for now – or think about how to define them first.)

What’s the most exotic function that you can study? Can you build a function with exactly one point of continuity? Exactly two? Exactly one hundred? Infinitely many points of continuity and infinitely many points of discontinuity? Exactly one point of discontinuity? Exactly two? Can you build a function that is continuous on some intervals and not on others? You should try to get a feeling for what is possible. You could try this for real-valued functions and functions from the complex numbers to the complex numbers.

Remind yourself (without looking at your notes!) what the definition of continuity is. What’s the negation? That is, what does it mean to say that $f$ is not continuous at a point $a$ ?

Preparation for Lecture 8

In Lemma 14 we considered various ways to combine functions to obtain another. What happens with composition of functions? If $f$ and $g$ are defined appropriately (what does that mean?) and are continuous at appropriate points (what does that mean?), is the composition continuous at an appropriate point?

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5 Responses to “Analysis I: Lecture 7”

KMacfarlane Says:
February 1, 2013 at 2:49 pm
Hi I have a small pedantic comment about our proof of lemma 14.

In the final lines of (iii) and (iv) I think you wrote a weak (<=) inequality on the board when it should be a strong (<) inequality.

This is needed to match the definition of continuity,
ie. we need |f(z) – f(a)| to be strictly less than epsilon, no matter how small epsilon is.
Zhixun Liang Says:
February 1, 2013 at 4:20 pm
Actually it doesn’t matter, as epsilon is arbitrary. As you said, no matter how small epsilon is.
cameron Says:
February 1, 2013 at 5:05 pm
Hi,
In part (iv) of Lemma 14 you insist that f(z) is not 0 for every z in E. Is this not overly restrictive? For example f(z)=0 at z=0 so we could not apply this Lemma so deduce g(z)=1/z is continuous at z=3, and yet it is. Is there some way we can alter this Lemma to make it more applicable? Maybe insist only f(z) is not zero for a small interval either side of the point of interest?
Thanks,
Cameron
theoremoftheweek Says:
February 1, 2013 at 7:10 pm
Cameron, this is a good question. It’s important to be careful about what the domain (E) of our function is. If 0 is in the domain, then we can’t define f by f(z) = 1/z — that function is not defined at the origin. (We could define a function from the reals to the reals by putting f(z) = 1/z for z non-zero and f(0) = 1, for example.) If you exclude 0, then the issue doesn’t arise.

Does that help? Do ask if you’re still unclear.
KMacfarlane Says:
February 3, 2013 at 9:34 am
Zhixun Liang, thanks for pointing that out – clearly we could take e to be epsilon/2 then clearly if we are <= e we are < epsilon.

Theorem of the week