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Challenge Yet another counterexample challenge!

  1. Apr 28, 2016 #1

    micromass

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    Well, the last thread of counterexamples was pretty fun. So why not do it again! Again, I present you a list with 10 mathematical statements. The only rub now is that only ##9## are false, thus one of the statements is true. Provide a counterexample to the false statements and a proof for the correct one!

    Rules:
    • For a counterexample to count, the answer must not only be correct, but a detailed argumentation must also be given as to why it is a counterexample.
    • Any use of outside sources is allowed, but do not look up the question directly. For example, it is ok to go check analysis books, but it is not allowed to google the exact question.
    • If you previously encountered this statement and remember the solution, then you cannot participate in this particular statement.
    • All mathematical methods are allowed.

    Here you go:

    1. SOLVED BY PeroK ##\mathbb{R}## is the disjoint countable union of closed intervals
    2. SOLVED BY mfb Any open set in ##\mathbb{R}^2## is the disjoint countable union of open balls, where an open ball is a set of the form ##B(\mathbf{a},r) = \{\mathbf{x}\in \mathbb{R}^2~\vert~(a_1 - x_1)^2 + (a_2 - x_2)^2 < r^2\}## for ##\mathbf{a}\in \mathbb{R}^2## and ##r>0##.
    3. SOLVED BY andrewkirk Let ##\pi:\mathbb{N}\times \mathbb{N}\rightarrow \mathbb{N}## be a bijection and let ##\sum_{n} a_n## be an absolutely convergent series in ##\mathbb{R}##. Then ##\sum_n a_n = \sum_{k=0}^\infty \sum_{l=0}^\infty a_{\pi(k,l)}##.
    4. SOLVED BY PeroK Every continuous bounded function ##\mathbb{R}\rightarrow \mathbb{R}## is uniformly continuous.
    5. SOLVED BY Samy_A Every function ##\mathbb{R}\rightarrow \mathbb{R}## is the derivative of some function.
    6. SOLVED BY PeroK If ##f:\mathbb{R}\rightarrow \mathbb{R}## is continuous and sends open sets to open sets, then it also sends closed sets to closed sets.
    7. SOLVED BY mfbThere is no monotonic function ##\mathbb{R}\rightarrow \mathbb{R}## whose set of discontinuities is ##\mathbb{Q}##.
    8. SOLVED BY Samy_A Every nonconstant function ##\mathbb{R}\rightarrow \mathbb{R}## that is periodic has a smallest period.
    9. SOLVED BY andrewkirk For any infinitely differentiable monotonic function ##f:\mathbb{R}\rightarrow \mathbb{R}## such that ##\lim_{x\rightarrow +\infty} f(x) = 0##, also holds that ##\lim_{x\rightarrow +\infty} f^\prime(x) = 0##.
    10. SOLVED BY andrewkirk Every subset of ##[0,1]## of measure zero is the set of points of discontinuity of a Riemann-integrable function ##[0,1]\rightarrow \mathbb{R}##.

    Thank you all for participating! I hope some of these statements were surprising to some of you and I hope some of you have fun with this! Don't hesitate to post any feedback in the thread!
     
    Last edited: May 10, 2016
  2. jcsd
  3. Apr 28, 2016 #2

    Samy_A

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    Define ##f:\mathbb{R}\rightarrow \mathbb{R}## as follows
    ##\left\{\begin{array}{l}
    x \in \mathbb Q : x \mapsto 1\\
    x \notin \mathbb Q : x \mapsto 2
    \end{array}\right.##
    Every rational number is a period, since for every rational number ##q##, ##x+q## will be rational if and only if ##x## is rational. That implies that for every rational number ##q##, ##\forall x \in \mathbb R: f(q+x)=f(x)##.
     
  4. Apr 28, 2016 #3

    Samy_A

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    That seems trivial. Did you mean continuous bounded function?
     
  5. Apr 28, 2016 #4

    micromass

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    Yes, sorry!
     
  6. Apr 28, 2016 #5

    Samy_A

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    I feel there must be an easier one, but this should work.
    The derivative of a function must be Lebesgue measurable.

    Proof: assume ##f=g'##
    That means that ##\displaystyle \forall x \in \mathbb R: \lim_{h \rightarrow 0} \frac{g(x+h)-g(x)}{h}=g'(x)=f(x)##.
    Define ##f_n:\mathbb{R}\rightarrow \mathbb{R}: x \mapsto \frac{g(x+1/n)-g(x)}{1/n}##
    Then ##\displaystyle \forall x \in \mathbb R: \lim_{n \rightarrow +\infty}f_n(x)=g'(x)=f(x)##.
    Since ##g## is differentiable, it is also Lebesgue measurable. Hence the functions ##f_n## are also Lebesgue measurable.
    ##f## being the pointwise limit of Lebesgue measurable function is itself Lebesgue measurable.

    So a counterexample is a function ##f## that is not Lebesgue measurable.
     
  7. Apr 28, 2016 #6

    Samy_A

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    An easier one:

    Define ##f:\mathbb{R}\rightarrow \mathbb{R}## as follows
    ##\left\{\begin{array}{l}
    x \in \mathbb Q : x \mapsto 1\\
    x \notin \mathbb Q : x \mapsto 0
    \end{array}\right.##

    Notice ##f=0## almost everywhere.

    If ##f=g'##, then ##\displaystyle \int_0^a g'(x)dx=g(a)-g(0)##. (EDIT: I wonder: for this I probably need to first prove that g' is measurable, so we need the proof in the previous post anyway.)
    But ##\displaystyle \int_0^a g'(x)dx =\int_0^a f(x)dx =0##.
    Hence ##g## is constant, and its derivative must be 0. We have reached a contradiction.
     
    Last edited: Apr 28, 2016
  8. Apr 28, 2016 #7

    PeroK

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    Number 4 seems quite easy. ##f(x) = sin(x^2)##

    Let ##x_n = (n\pi)^{1/2}## and ##y_n = [(n + 1/2)\pi]^{1/2}##

    ##\forall n \ \ f(x_n) = sin(n\pi) = 0## and ## f(y_n) = sin(n\pi + \pi/2) = \pm 1##

    hence ##|f(x_n) - f(y_n)| = 1##

    But ##\lim_{n \rightarrow \infty}|x_n - y_n| = 0## (*)

    Therefore, ##\forall \delta > 0## we can choose ##n## such that ##|x_n - y_n| < \delta## but ##|f(x_n) - f(y_n)| = 1##

    Hence ##f## is not uniformly continuous.

    (*) ##|x_n - y_n|(x_n + y_n) = \pi/2##

    ##|x_n - y_n| = \frac{\pi}{2(x_n + y_n)} < \frac{\pi}{2\sqrt{n\pi}}##
     
  9. Apr 28, 2016 #8

    PeroK

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    Does that make this equivalent to the axiom of choice?
     
  10. Apr 28, 2016 #9

    Samy_A

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    Think so (for the existence of not Lebesgue measurable functions). But you don't need the axiom of choice to prove that a derivative is Lebesgue measurable. So my second example probably works without invoking the axiom of choice.
     
  11. Apr 28, 2016 #10

    micromass

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    Here is another proof: https://en.wikipedia.org/wiki/Darboux's_theorem_(analysis)
     
  12. Apr 28, 2016 #11

    micromass

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    This is a very interesting remark, and highly depends on what kind of integral you're looking at. Let's first look at the classical fundamental theorem of calculus (quoted from wiki):

    So we see in your counterexample, that you will first need to prove that ##g^\prime(x)## is Riemann integrable. This is obviously not the case. The classical fundamental theorem is not strong enough to let you do what you want.

    Luckily, there is an extension of the Riemann integral, called the Henstock integral. This integral has a very similar definition to the Riemann integral, but has an improved fundamental theorem. Its theory is explained in Bartle "A modern theory of integration" https://www.amazon.com/Modern-Integration-Graduate-Studies-Mathematics/dp/0821808451 Let me quote the relevant part

    (So this fundamental theorem actually proves Henstock integrability of ##f##. No further assumptions are necessary other than it being a derivative. Not even measurability. (Note that Henstock integrability implies measurable though). So this theorem does exactly what you want. Note that this isn't even the most general theorem of this kind, you can get away with a lot less than what the quoted part assumed, all of that is in Bartle.

    Now I could be very justified to put in a rant on why Henstock integrals are not at all taught in schools. They are extremely easy to define, and are more general than Lebesgue integrals (for functions ##\mathbb{R}\rightarrow \mathbb{R}##), and they have way better properties than the silly Riemann integral. Sadly, they are ignored by the vast majority of books (one notable exception is DePree Swartz https://www.amazon.com/Introduction-Real-Analysis-John-DePree/dp/0471853917)
     
  13. Apr 28, 2016 #12

    PeroK

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    I assume for number 1 the statement should be "not" and then the counterexample is R given as a disjoint countable union of closed intervals. I'll do non-negative R.

    The first set of intervals are ##[0, 1], [2, 3], [4, 5] \dots##

    The second set of intervals are ##[1\frac{1}{4}, 1\frac{3}{4}], [3\frac{1}{4}, 3\frac{3}{4}] \dots##

    The third set of intervals are ##[1\frac{1}{16}, 1\frac{3}{16}], [1\frac{13}{16}, 1\frac{15}{16}], [3\frac{1}{16}, 3\frac{3}{16}], [3\frac{13}{16}, 1\frac{15}{16}] \dots##

    Etc.
     
  14. Apr 28, 2016 #13

    jbriggs444

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    But.

    It seems that you have provided at least the core of a proof for the correct one.
     
  15. Apr 28, 2016 #14

    jbriggs444

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    The idea is that one wants to come up with a kind of smoothed step function where the steps get smaller and smaller but retain a similar slope. If one had as a building block a monotone increasing function that is continuous and infinitely differentiable on the unit interval [0,1] with all derivatives equal to zero at both endpoints then one could use that to construct the counter example by splining such building blocks together. Unfortunately, my toolbag only contains a function with the required properties at one endpoint.

    Wikipedia calls these "smooth transfer functions". And I think at this point I think I've been spoiled too badly to compete.
     
  16. Apr 28, 2016 #15

    micromass

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    So, the easiest is to write this in base ##4## notation. Let me focus on the cover of ##(1,2)## which according to you is:

    [tex][0.1, 0.3][/tex]
    [tex]0.01, 0.03],~[0.31, 0.33][/tex]
    [tex][0.001, 0.003],~[0.031, 0.033],~[0.0301, 0.303],~[0.331, 0.333][/tex]
    And so on.

    So the first step covers points of the form ##0.1xxxx...## and ##0.2xxx...## (assume terminating "decimals" are forbidden). No point with ##1## and ##2## in the first place remains in the remainder.
    The second step covers points of the form ##0.01xxx...## and ##0.02xxx...## and ##0.31xxx...## and ##0.32xxx##. Removing this, we see that no point with ##1## and ##2## in the first two places remain.

    All in all, in the ##n##th step, we see that no point with ##1## and ##2## in the first ##n## decimals places remain in the remainder.

    So what your collection of closed intervals covers are all points with ##1## and ##2## in some decimal place. But from this description it is easy to see that the point ##0.030303030303...## is not covered by your collection of intervals. So your counterexample is not valid.
     
  17. Apr 28, 2016 #16

    micromass

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    Don't worry, you can still compete with this question if you want to. After all, checking wikipedia or textbooks for specific subjects is allowed, as long as you don't find a direct answer to the question.
     
  18. Apr 28, 2016 #17

    Samy_A

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    Aahh, that's the one!
    I knew there was some well known property that all derivatives had, but couldn't remember which one.
    Very cool. Henstock integral: something to look at.
     
  19. Apr 28, 2016 #18

    micromass

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    Well, we have detailed some of its properties already: derivatives are Lebesgue measurable and satisfy the conclusion of the intermediate value theorem. One could ask whether this completely characterizes derivatives. But that's for a next set of counterexamples (the answer is no).
     
  20. Apr 28, 2016 #19

    fresh_42

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    I define ##f(x) = x^{-2}## for ##x\neq 0## and ##f(0)=0##.
    My "fraud (?)" is this: I take the discrete topology on both ##\mathbb{R}##. Then ##f## is clearly open.
    With ##f^{-1}(\{x\}) = \{\frac{1}{\sqrt{x}}\}## for ##x > 0##, ##f^{-1}(\{0\})=\{0\}## and ##f^{-1}(\{x\})=∅## for ##x < 0## ##f## is also continuous.
    Now ##f(\mathbb{R} - \{0\}) = \mathbb{R^+} = ∪_{x>0} \{x\}## maps a closed set onto an open.
     
    Last edited: Apr 28, 2016
  21. Apr 28, 2016 #20

    micromass

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    OK, but your map still takes closed sets to closed sets.

    (And I kind of meant ##\mathbb{R}## to have the Euclidean topology)
     
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