# Extra Maths: Analysis, Accumulation

Here are three fun Analysis questions, really aimed at my students who are studying their first/second course on Real Analysis, but hopefully it’s of wider interest as well.

We begin at the beginning, with a couple of definitions.

Definition. A function $f:\mathbb{R}\longrightarrow\mathbb{R}$ has the IVT Property if, for all real numbers $a,b,x$ with $a and $\min\{f(a),f(b)\} there exists $\xi\in[a,b]$ such that $f(\xi)=x$.

So, using this terminology, the Intermediate Value Theorem is the statement that a continuous function $\mathbb{R}\longrightarrow\mathbb{R}$ has the IVT Property.

Definition. Let $f:\mathbb{R}\longrightarrow\mathbb{R}$ and let $a,l\in\mathbb{R}$. We say that $l$ is a weak limit of $f$ at $i$ if
for all $\varepsilon>0$ there exists $b\in(a-\varepsilon,a+\varepsilon)\setminus\{a\}$ such that $f(b)\in(l-\varepsilon,l+\varepsilon)$. Naturally, we say that $f$ has a weak limit somewhere if there exists $a,l\in\mathbb{R}$ such that $l$ is a weak-limit of $f$ at $a$.

We say that $l\in\mathbb{R}$ is an accumulation point of a real sequence $(a_{n})$ if for all $\varepsilon>0$ there is an infinite set $N\subseteq\mathbb{N}$ such that for all $n\geq N$ we have $|a_{n}-l|<\varepsilon$.

The Questions

Question 1.
Show the existence or non-existence of a function $f:\mathbb{R}\longrightarrow\mathbb{R}$ which has the IVT Property but which is nowhere continuous.

Question 2.
Prove or disprove the claim that every function $f:\mathbb{R}\longrightarrow\mathbb{R}$ has a weak limit somewhere.

Question 3.
Find a real sequence $(a_{n})$ such that every real number $l\in\mathbb{R}$ is an accumulation point of $(a_{n})$.

Where should we start? Perhaps my favourite of the three is the first one: to show the existence or non-existence of a function with the IVT property but which is nowhere continuous. There are several approaches we might take, beginning with the following lemma.

Lemma 1. Suppose that a function $f:\mathbb{R}\longrightarrow\mathbb{R}$ has the property that, for all $a,b$ with $a, the restriction of $f$ to the interval $(a,b)$ is surjective onto $\mathbb{R}$. Then $f$ has the IVT property.

Hopefully the proof of this lemma is quite clear: the IVT property asks, of such a function restricted to such an interval, that it attain certain values. So if $f$ attains all real values, then the property is evidently satisfied.

Okay…. So what?

Lemma 2. Suppose that a function $f:\mathbb{R}\longrightarrow\mathbb{R}$ has the property that, for all $a,b$ with $a, the restriction of $f$ to the interval $(a,b)$ is surjective onto $\mathbb{R}$. Then $f$ is continuous nowhere.

Proof. Let $a\in\mathbb{R}$ be given, and consider, for example, $\varepsilon=1$. For any $\delta>0$, by assumption $f$ restricted to the open interval $(a-\delta,a+\delta)$ is surjective onto $\mathbb{R}$. In particular, the image under $f$ of $(a-\delta,a+\delta)$ is not a subset of $(f(a)-1,f(a)+1)$. $\square$

Ah-ha! Now it’s clear why we’re interested in these sorts of function. Such a `locally surjective function‘ would be a nowhere continuous function which nevertheless has the IVT property. The big question now is: does any locally surjective function exist?

Well, YES. Such functions do exist, and there are several ways to find them. First we give a slightly abstract construction, which might seem a little unsatisfactory.

Method 1. Recall that the cardinality of $\mathbb{R}$ is $2^{\aleph_{0}}$, and the cardinality of $\mathbb{R}/\mathbb{Q}$ is also $2^{\aleph_{0}}$. We may choose a bijection

$F:\mathbb{R}/\mathbb{Q}\longrightarrow\mathbb{R}$.

Also, let $\phi:\mathbb{R}\longrightarrow\mathbb{R}/\mathbb{Q}$ be the natural map sending each real number $r$ to its coset modulo $\mathbb{Q}$, namely $r+\mathbb{Q}$. Note that $\phi$ is surjective. Thus, the composition $F\circ\phi:\mathbb{R}\longrightarrow\mathbb{R}$ is also a surjection. Indeed, by the density of the rationals in the reals, each proper open interval $(a,b)$ contains a representative of each coset of $\mathbb{Q}$ in $\mathbb{R}$. That is, for each $r\in\mathbb{R}$, there exists $q\in\mathbb{Q}$ such that $r+q\in(a,b)$. A consequence of this is that the restriction to $(a,b)$ of $\phi$ is still surjective onto $\mathbb{R}/\mathbb{Q}$. Finally, the restriction of $F\circ\phi$ to $(a,b)$ is still surjective onto $\mathbb{R}$, which shows that $F\circ\phi$ is locally surjective, as required.$\square$

Wonderful, that was easy. But it wasn’t particularly constructive. Perhaps it felt a little unsatisfactory to rely on the cardinalities of these sets. Is there a better way?

That’s where I’ll leave it for the moment. To be continued!

In the meantime, try to construct a locally surjective function by using the decimal expansion of each real number.