Gamelin and Green, Introduction To Topology Ch. 1 section 3 problem 10.

Let $f$ be a real-valued function on $\mathbb{R}$, the real numbers. Show that there exist $M \gt 0$ and a nonempty open subset $U$ of $\mathbb{R} $ such that for any $s \in U$ , there is a sequence $\{s_n\}$ satisfying $s_n \rightarrow s$ and $|f(s_n)| \le M, n \ge 1$.

I intuitively get this. Every point in an open set has a sequence converging to it, so this is just saying if any $s_n$ are "too big" I can find another $s'_n$ close by that isn't "too big". ($s$, itself, can be as big as it wants.)

This is obviously true if $f$ is continuous or even just bounded on any open interval. (Just let U be the image $f(A)$ of any open interval, $A$, and let $M = \sup f(A)$). For this not to be true, $f$ needs to be unbounded on every possible open set. Not merely unbounded which is easily possible but.... If $V_M =\{x : f(x) \ge M\}$ will always always be open no matter how large the $M$. Which seems intuitively impossible. But, I don't have it.

At least I don't with concepts of the first 15 pages of the book. (Metric spaces, closed and open sets, convergent sequences, reals having least upper bound property, but not compactness; it's a pretty dense book.)

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    $\begingroup$ AnatolyVorobey did provide an acceptable answer but as f need not be continuous nor bounded on any open interval, I don't see anything "obvious" about it. $\endgroup$ – fleablood Sep 25 '16 at 18:43

New approach!

$x$ is called good for $M$ if there's a sequence $\{x_i\}$ tending to $x$ with all its $f$-values $\le M$.

Clearly every point $x$ is good for some $M$, for example for $f(x)$.

It follows that if we define $G_M = \{x \in R: x \text{ good for } M\}$, then the countable union $G_1\cup G_2 \cup G_3 \cdots$ is all of $\mathbb{R}$.

If only some $G_M$ were open, that would have settled the problem. But actually life isn't that easy. Actually each $G_M$ is closed. Let's prove this. Take $\{x_i\}$ a sequence of points in some $G_M$ tending to $x$. Use the good sequences tending to every $x_i$ and combine them to build a good sequence tending to $x$, therefore $x \in G_M$ (details omitted).

OK, so if $G_M$ is closed, then let's take its interior. If it's nonempty, that will be the open set we need.

But maybe all the closed sets $G_M$, for every $M$, have empty interiors. We will prove that can't be the case using the Baire category theorem.

If all $G_i$, $i=1,2,3,\dots$ have empty interiors, then their complements $U_i = R\setminus G_i$ are dense open sets. Since the union of $G_i$ is $\mathbb{R}$, the intersection of $U_i$'s is empty. But by the Baire category theorem this intersection is a dense (so certainly nonempty) set. Contradiction.

Therefore at least one $G_i$ will have a nonempty interior, and its interior will be the open set we need.

  • $\begingroup$ "Clearly every point x is good for some M, for example for f(x)" I don't see that this is true. In fact, I'm sure it is false. I just don't believe that all points can be bad and at least one point must be good. But I can't prove this. $\endgroup$ – fleablood Oct 12 '15 at 21:53
  • $\begingroup$ Let M_e_x but the inf of values of |f(y)|, y != x y in e-ball of x. For smaller e M_e_x are larger. The sequence of M_e_x need not be bounded. If not, the x is a bad point. All I need to do is show that somewhere there exists a good point x where the sequences of M_e_x are bounded and have a sup. An open neighborhood around a good x woould suffice. But I don't know that any good x's exist. $\endgroup$ – fleablood Oct 12 '15 at 21:58
  • $\begingroup$ Say |f(x)| = M but for all the y d(y,x) < e |f(y)| > M. Then x is not good for M. $\endgroup$ – fleablood Oct 12 '15 at 22:01
  • $\begingroup$ No, you're missing the point that a constant sequence $\{x,x,x,x,x,...\}$ also works, if $f(x) \le M$. There's no requirement that the points in the sequence that converges to $x$ be distinct from $x$. $\endgroup$ – AnatolyVorobey Oct 12 '15 at 22:10
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    $\begingroup$ Ah, I see. x isn't going to be the s we want. Ah, Baire! I forgot about that. Okay, I think this will work! Let me chew on it. $\endgroup$ – fleablood Oct 12 '15 at 22:14

EDIT: the approach below doesn't work.

Since you know absolutely nothing about continuity or any kind of nice behavior of $f$, it seems fruitless to try and correlate its behavior with open sets and other such topological objects (what does $f$ care between an open set and some other random continuum-sized subset of $R$?).

But, here's a nice thing, you're only asked to evaluate $f$ on the $s_n$ that you provide - not on $s$. This is important! If we always choose $s_n$ from a very limited stock of points $S$, we only need to care about values of $f$ on those points and nothing else. But as long as $S$ is continuum-sized, $f$ can be just as horrible on $S$ as it is on the whole of $R$.

Hmm, do we have a less-than-continuum-sized subset $S$ of $R$ such that every real has a sequence in $S$ convergent to it? Right!

(I advise to stop reading here and try continue on your own for a while, come back if it doesn't work out).

OK, now take any $x \in R$ and $M > 0$. Maybe $x$ has a sequence of rationals converging to it, in which we can find a subsequence which has all its $f$-values below $M$. In this case let's say that $x$ is a good point for $M$.

Is it possible that there's no pair $(x,M)$ such that $x$ is good for $M$? No; if $x$ itself is rational, it is trivially good for $f(x)$.

OK, so for some $M$ there are good points. Maybe all the good points for $M$ together make an open set? Hint: look at the bad points for $M$ and try to show it's a closed set.

  • $\begingroup$ " look at the bad points for M and try to show it's a closed set." That is precisely the problem I have not been able to figure out. I hadn't tried restricting to rationals though... Not seeing it right now. I figure if M gets large enough the bad points "thin" but I can't seem to prove will never cease to be adherent to a bad point. I figure this is a "they can't all be above average" issue but I'm not seeing it. $\endgroup$ – fleablood Oct 11 '15 at 23:55
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    $\begingroup$ To show that the bad points form a closed set, it's enough to show that if $x_i$ is a sequence of bad points which tends to $x$, then $x$ is bad too. Say that a point $q$ is "large" if $f(q) > M$. Then each $x_i$ being bad means it only has a finite number of non-large points, meaning you can find large rational points arbitrarily close to each $x_i$. Combine those to build a sequence of large rational points tending to $x$. (I think that's not enough under my definition of "good point", but if I modify it to require every rational sequence to have a behaving subsequence, it works out) $\endgroup$ – AnatolyVorobey Oct 12 '15 at 0:02
  • $\begingroup$ OK, this doesn't work because the new definition of "good point" makes it false that there has to be any good point for any M. Counterexample: order all rationals and assign $f(q)=$ $q$'s number in the order. $\endgroup$ – AnatolyVorobey Oct 12 '15 at 0:20
  • $\begingroup$ "being bad means it only has a finite number of non-large points" that's what I can't figure out. Suppose s is surrounded by large points. Every neighborhood will have an M value we could use to judge some point to be non-large. But a smaller neighborhood may require larger M. Let $M_q_e$ be an M such that the e-ball around q has a point smaller than M. There's no guarantee these M's converge. It seems absurd that these M's would diverge for all s in R, but I can't figure out how to argue that. $\endgroup$ – fleablood Oct 12 '15 at 0:22

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