3
$\begingroup$

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.)

$\endgroup$
  • 1
    $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ "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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ " 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.