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I was reading (self study) the book by Thomson Bruckner and Bruckner and ran into one of the exercises that I think I have the proof for by not sure it is right. The statement of the problem goes as follows.

A function $f:\mathbb{R} \mapsto \mathbb{R}$ is said to be bounded at a point $x_o$ provided there are positive number $\epsilon$ and $M$ so that $|f(x)| < M$. for all $x \in (x_o - \epsilon, x_o + \epsilon)$. Show that the set of points at which a function is bounded is open. Let $E$ be an arbitrary closed set. Is it possible to construct a function $f:\mathbb{R} \mapsto \mathbb{R}$ so that the set of points at which f is not bounded is precisely the set E?

My Proof of the first section of the problem was as follows. Let $B_f = \{x : |f(x)| < M, M > 0 \}$ the set of all points where $f$ is bounded (note $M = sup(\{M_x : x \in \mathbb{R}, |f(x)| < M_x\})$ over each $x$, This $sup(\{M_x : x \in \mathbb{R}, |f(x)| < M_x\})$ is bounded and exists, since each $M_x$ is bounded). Then if for every $x \in B_f$ we have by our definition of bounded function at a point, $\forall x \in B_f \ \exists \epsilon > 0 \text{ and } \forall x \in (x - \epsilon, x + \epsilon)$ we have $|f(x)| < M_x$, but this would mean $(x - \epsilon, x + \epsilon) \subset B_f$ which means that $x$ is a interior point of $B_f$. The conclusion follows that $B_f$ is open.

I have no clue how to construct the function for the second half of the question.

I was wondering if my proof is right? If someone can provide an example for the second part of the question I would be extremely thankful.

Ramesh Kadambi
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1 Answers1

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Your proof for the first part of the question has some problems.... Note that your set $B_f$ actually depends on the particular $M>0$ which is use to define it, but later you lose track of this dependence when you argue that $x$ is an interior point of $B_f$ using $M_x >0$, which may turn out to have $M_x > M$!

For the second half of the question, the answer is that, yes, for any closed set $E \subseteq \mathbb{R}$, there is a function $f:\mathbb{R} \to \mathbb{R}$ which blows up exactly on $E$. A good start is to define $$ \begin{align*} f(x) = \frac{1}{\mathrm{distance}(x,E)} && \text{ for all } x \in \mathbb{R} \setminus E \end{align*}$$ This partial definition guarantees that

  • $f$ is bounded at every point in $\mathbb{R} \setminus E$, and
  • $f$ is unbounded at every boundary point of $E$.

We need to finish defining $f$ so that it is also unbounded on the interior of $E$. For this, we clearly need a lemma like:

Lemma: Let $U$ be any open subset of $\mathbb{R}$. Then, there is a function $U \to \mathbb{R}$ which is unbounded at every point of $U$.

I'm sure there are various ways to prove this lemma. The first thing which came to my mind is the following argument

Proof: Let $x_1,x_2,x_3,\ldots$ be real numbers, chosen such that the cosets $x_n + \mathbb{Q}$ are pairwise disjoint. Define $g : U \to \mathbb{R}$ by $$g(x) = \begin{cases} \frac{1}{n} && \text{ if } x \in x_n +\mathbb{Q} \text{ for some } n \\ 0 && \text{ if } x \notin x_n + \mathbb{Q} \text{ for all } n \\ \end{cases}$$ Since each coset $x_n + \mathbb{Q}$ has dense intersection with $U$, $g$ blows up at each point of $U$. $\square$

Anyway, however this lemma is proved, we can find a function $g:\mathrm{interior}(E) \to \mathbb{R}$ which is unbounded at each point of $\mathrm{interior}(E)$ and use it to finish off our definition of $f$: $$f(x) = \begin{cases} \frac{1}{\mathrm{distance}(x,E)} && \text{ if } x \notin E \\ g(x) && \text{ if } x \in \mathrm{interior}(E) \\ 0 && \text{ if } x \in \mathrm{boundary}(E) \\ \end{cases}$$

Mike F
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  • I guess I missed out on the note. The point is that we can choose an M such that it is larger than each $M_x$. Thanks again for the second part. That is indeed clever. – Ramesh Kadambi Nov 03 '16 at 17:18
  • Mike F: Could you check my modifications to the proof and confirm that it is correct. – Ramesh Kadambi Nov 03 '16 at 19:34
  • @rk540: It is still not correct. That supremum need not be finite. Consider the function $f(x) = x$. There is no global bound on $f$, but $f$ is bounded (locally) at every point. – Mike F Nov 03 '16 at 20:29
  • Mike F: I see your point, on thinking further, I am now thinking $B_f$ should be defined as set of all points where $f$ is locally bounded. That would mean by definition $B_f$ is a union of all the open intervals $(x-\epsilon, x + \epsilon)$ and thus an open set? I did not think the proof was that simple against my first instincts. If not you can show that $(x-\epsilon, x+\epsilon) \subset B_f \ \forall x \in B_f$ by definition and hence every $x$ is an interior point. (I am not able to do @ your name not sure why) – Ramesh Kadambi Nov 04 '16 at 12:16
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    That's mostly correct, although if I was being picky I could still complain a little. Let $B_f$ be the set of points where $f$ is bounded, according to the definition in the problem statement. Thus, $x_0 \in B_f$ iff $f$ is bounded on $(x_0-\epsilon,x_0+\epsilon)$ for some $\epsilon > 0$. OK, but why does every point $x \in (x_0 - \epsilon,x_0+\epsilon)$ also belong to $B_f$? The point I'm trying to make (a rather minor one) is that, to show $x \in B_f$, one is supposed to check that $f$ is bounded on an interval centred at $x$, but the interval $(x_0-\epsilon,x_0+\epsilon)$ may not be. – Mike F Nov 04 '16 at 13:05
  • Thank you very much for the input and corrections. I do understand that i need to prove that each $x$ in the $x_o - \epsilon, x_o + \epsilon) \in B_f$, Will think about and see why that is so. – Ramesh Kadambi Nov 04 '16 at 15:06
  • BTW for the second part of the question a function that is not bounded on a closed set. Would $\frac{1}{1 - \mathbb{1}_E}$ work ? Where E is a closed set and $\mathbb{1}_E$ is a characteristic function of the set. If not why? – Ramesh Kadambi Nov 04 '16 at 17:17
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    That function is not defined on $E$ (division by zero). Unless you want it to take the value $\infty$ on $E$? Since you wrote $f:\mathbb{R} \to \mathbb{R}$ in your post, I assumed you only wanted real-valued functions. But sure, if you are OK with infinite values, then by all means take something like what you wrote, or even $f(x)=\begin{cases} \infty & \text{ if } x \in E \ 0 & \text{ otherwise} \ \end{cases}$. – Mike F Nov 04 '16 at 20:50
  • Mike F: Thank you. Yes I was looking for $f:\mathbb{R} \mapsto \mathbb{R}$. So I get it, The subtlety was missed. I appreciate the explanation. Thank you very much again. – Ramesh Kadambi Nov 06 '16 at 11:19
  • Mike F: I was rereading your answer and to answer your question as to why every $x \in (x_o - \epsilon, x_o + \epsilon)$ is also $\in B_f$, every point $x$ is an interior point. Given that $\exists \ M_{x_o}$ for every $x \in (x_o - \epsilon, x_o + \epsilon)$ the same $M_{x_o}$ holds for every $x$ and an interval centered around $x$ with an $(x - \epsilon_x, x + \epsilon_x)$. Essentially the fact that $f$ is locally bounded on an open interval means that it is true for every point in the interval as all points belong to the interior. – Ramesh Kadambi Dec 21 '16 at 14:19