# A-noncompact, Does there always exist a continuous function $f: A \to \mathbb R$ which is bounded but does not assume extreme values?

It's well known that if $A \subset \mathbb R$ is compact then every continuous function $f:A \to \mathbb R$ is bounded and assume extreme values .So the obvious question is:

Given any non compact set $A \subset \mathbb R$ does there always exist a continuous function $f: A \to \mathbb R$ which is bounded but does not assume extreme values?

Any ideas for constructing such function?

• Do you want the function to assume neither max nor min, or not to assume both max and min? – Vim Aug 23 '15 at 10:36
• @Vim I'm interested in a function which does not assume neither max nor min. – Arpit Kansal Aug 23 '15 at 16:21
• If $A$ is allowed to be an arbitrary topological space, not necessarily a subset of $\mathbb{R}$, then we can take $A = \omega_1$ to be the first uncountable ordinal space. This space is not compact but every continuous real-valued function assumes extreme values. – Nate Eldredge Aug 23 '15 at 19:33
• @NateEldredge Thank you,Regards – Arpit Kansal Aug 23 '15 at 19:48

Let $\langle X,d\rangle$ be any non-compact metric space, not necessarily a subspace of $\Bbb R$. Then $X$ is not countably compact, so $X$ has a countably infinite closed discrete subset $D=\{x_n:n\in\Bbb N\}$. Define

$$f:D\to\Bbb R:x_n\mapsto\begin{cases} 1-2^{-n},&\text{if }n\text{ is even}\\ -1+2^{-n},&\text{if }n\text{ is odd}\;. \end{cases}$$

By the Tietze extension theorem there is a continuous $g:X\to[-1,1]$ such that $g\upharpoonright D=f$. Every metric space is perfectly normal, so $D$ is a zero set, and there is a continuous $\varphi:X\to[0,1]$ such that $\varphi(x)=0$ if and only if $x\in D$. Let $\psi:X\to[0,1]:x\mapsto 1-\varphi(x)$, and let

$$h:X\to\Bbb R:x\mapsto g(x)\psi(x)\;;$$

then $h$ is continuous, $\inf_{x\in X}h(x)=\inf_{x\in D}h(x)=-1$, $\sup_{x\in X}h(x)=\sup_{x\in D}h(x)=1$, and $h[X]\subseteq(-1,1)$, so $h$ attains neither its supremum nor its infimum.

Added: I’ve been asked to say something about how I thought of this construction. That’s a bit difficult, since the pieces were all just lying about in my head and came together without much effort on my part, but I’ll try. I know that countable compactness and compactness are equivalent for metric spaces, so I know right away that a non-compact metric space has a countably infinite closed discrete subset $D$. Because $D$ is discrete, any function from it to $\Bbb R$ will be continuous so I’ll begin by defining a bounded function from $D$ to $\Bbb R$ that doesn’t attain its supremum or infimum. There are may such; I chose to use the function $f$ above, but any other would have worked equally well.

This particular function $f$ takes values in $(-1,1)$ and has $-1$ and $1$ as its unattained extrema. If I can extend $f$ to a continuous $h:X\to(-1,1)$, I'll be done, so what do I know about extending continuous functions? Since $D$ is closed, the Tietze extension theorem applies, and it almost does what I want: it guarantees the existence of a continuous $g:X\to\Bbb R$ with the same extrema as $f$, which implies that $g[X]\subseteq[-1,1]$. Now I just have to make sure not to hit $-1$ or $1$. I can’t do that with $g$ itself, but if I can find a continuous function $\psi:X\to[0,1]$ that is $1$ only on $D$, the product $g\psi$ is guaranteed so map $X\setminus D$ into $(-1,1)$, and I’ll be home free. That requires that $D$ be a zero set in $X$, but I know that that’s true of every closed set in a metric space, so I’m done.

By the way, here’s an example to show that the result does not hold for arbitrary non-compact spaces. The ordinal space $\omega_1$ of all countable ordinals with the order topology is a standard example of a countably compact space that isn’t compact, so in particular it’s not metric. It happens that every continuous $f:\omega_1\to\Bbb R$ is eventually constant: there is an $\alpha_f<\omega_1$ such that $f(\alpha)=f(\alpha_f)$ whenever $\alpha_f\le\alpha<\omega_1$. The set $[0,\alpha_f]$ is a compact subset of $\omega_1$, so $f\upharpoonright[0,\alpha_f]$ attains its extrema, and $f(\alpha)=f9\alpha_f)$ for $\alpha\in\omega_1\setminus[0,\alpha_f]$, so $f$ has the same extrema as $f\upharpoonright[0,\alpha_f]$ and obviously attains them.

• Thank you very much,I need some time to digest it properly,Regards, – Arpit Kansal Aug 23 '15 at 19:24
• @Arpit: You’re welcome; by all means take your time. – Brian M. Scott Aug 23 '15 at 19:25
• Sorry..I "understand" the solution but I don't see how do you think of such construction.Can you please say some lines about construction? – Arpit Kansal Aug 23 '15 at 19:43
• @ArpitKansal Maybe try to apply it to $\mathbb{R}$ itself and to $(0,1)$, by explicitly producing the Tietze extension. – Ian Aug 23 '15 at 21:36
• @BrianM.Scott It's very clear now,Thanks a lot.Best Regards, – Arpit Kansal Aug 25 '15 at 8:35

A set $A \subset \mathbb{R}$ is compact iff is is bounded and closed. So if a set $A$ is not compact, it must be unbounded or not closed [or both].

If $A$ is not closed, there is at least one point $b \in A^c$ which is a limit point of A. Wlog $b = 0$ and we can consider the function $$x \mapsto \begin{cases} \sin(\tfrac{1}{x}) (x + 1)^{-1} & x > 0 \\ 0 & x < 0 \end{cases}.$$

If $A$ is closed, it must be unbounded. If it is unbounded towards the top and the bottom, you can consider $x \mapsto \arctan(x)$. If it is not unbounded in both directions, we can wlog assume that it is bounded from below by $0$. Now choose a sequence of points $0 = x_0 < x_1 < x_2 < \ldots$ in $A$ with $x_n \to \infty$ and define the function $g: \mathbb{R}^+ \to [-1, 1]$ by $g(x_i) = (-1)^i$ and linearize it inbetween the $x_i$. The function $x \mapsto \arctan(x) g(x)$ is bounded, but it attains neither its maximum nor minimum.

If $A$ is not closed, it has a limit point $b \in A^c$. Wlog $b = 0$ and there exists a sequence $x_1 > x_2 > \ldots$ in $A$ that converges to $0$. Repeat the construction of $g$ from the unbounded case and consider the function $$x \mapsto \begin{cases} g(x) (x + 1)^{-1} & x > 0 \\ 0 & x < 0 \end{cases}$$

• That first function will not work for $A=\{1/(2n\pi) : n \in \mathbb N \}.$ – zhw. Aug 23 '15 at 21:18
• @zhw. You are right, I will correct it. – Dominik Aug 23 '15 at 21:26

The following will work in any $\mathbb {R}^k.$ Let $g$ be the continuous function on $\mathbb {R}^k$ defined by $g(x) = 1-|x|,|x|\le 1, g(x) =0, |x|>1.$ I'll use $g$ in the constructions below.

If $A\subset \mathbb {R}^k$ is not closed, then there is a sequence of distinct points $(a_n)$ in $A$ such that $a_n \to a \not \in A.$ We can then find pairwise disjoint closed balls $\overline {B(a_n,r_n)}, n=1,2,\dots ,$ that do not contain $a.$ For $x\in \mathbb {R}^k,$ define

$$\tag 1 f(x) = \sum_{n=1}^{\infty}(-1)^n(1-1/n)g(2(x-a_n)/r_n).$$

Note the $n$th summand has support in $\overline {B(a_n,r_n/2)};$ this insures the supports stay well away from each other. The function $f$ is continuous on $\mathbb {R}^k\setminus \{a\},$ with range contained in $(-1,1).$ Note that $f(a_n) = (1-1/n)$ if $n$ is even, and $f(a_n) = -(1-1/n)$ if $n$ is odd. Thus $f$ has no extreme values in $A.$

If $A$ is not bounded, then we do the same thing. This time it's easier to think about, since now $A$ contains a widely spaced sequence $(a_n)$ such that $|a_n| \to \infty.$ A sum as in $(1)$ will then yield a continuous function on $\mathbb {R}^k$ with range contained in $(-1,1)$ that has no extreme values in $A.$

*Edited from the original to correct a few errors and address the boundedness issue, which I somehow missed the first time around.

• The question was whether there is a function that attains neither its maximum nor its minimum [see also the comments in the first post]. – Dominik Aug 23 '15 at 19:53
• You're right, thanks. I'll leave this for a bit and think about it. If I can't think of anything decent, I'll delete the answer. – zhw. Aug 23 '15 at 20:52

Let $A \subset \mathbb{R}$ be a noncompact set. Consider the function $f: A \to \mathbb{R}$ defined by $f(x) = \frac{1}{1+e^{-x}}$, the logistic function. Note that $f(x) \in (0,1)$ for all $x\in \mathbb{R}$ and therefore $f$ is bounded regardless of $A$. However, $f$ never assumes its extreme values on any noncompact set.

• This function attains its maximum on $(-\infty, a]$ and its minimum on $[a, \infty)$. – Dominik Aug 23 '15 at 9:53
• @Dominik Hmm good point actually. ^^ – molarmass Aug 23 '15 at 9:55

Take $f(x)=\frac{1}{1+x^2}$, $x\in (1,3)$ then clearly $f$ is bounded and $f$ does not assumes any extreme value.

• who is make negative repu.? and why ? this function is bounded and no max nor min in the interval – Mohammad W. Alomari Aug 23 '15 at 12:19
• The question was whether one can find such a function for every noncompact set. – Dominik Aug 23 '15 at 15:49
• No, the question is: Does there exists a continuous function defined on a non-compact set which is always has no max nor min. – Mohammad W. Alomari Aug 23 '15 at 17:22
• If this was really the question, the word "always" would be pointless. Also, the question starts with "Given any non compact set". – Dominik Aug 23 '15 at 17:24