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<b and \min\{f(a),f(b)\}<x<\max\{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<b, 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<b, 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.

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