Your most counterintuitive yet simple math problem

  • #51
2,112
18
A feminist claim: As is known, women are more loyal to their partners than men are. So on average, men have more sexual relationships than women have.

:wink:
 
  • #52
534
1
In general, for any function [itex]f \colon \mathbb{R} \to \mathbb{R}[/itex] whose value is known everywhere except [itex]x_0[/itex], the remaining value could be anything. However, if the function was continuous, then there is only one possibility for [itex]f(x_0)[/itex]. This generalizes by the theorem below.

Theorem. Let [itex]X[/itex] and [itex]Y[/itex] be topological spaces, with [itex]Y[/itex] Hausdorff. Let [itex]x_0[/itex] be a limit point of [itex]X[/itex], and let [itex]f \colon X \to Y[/itex] and [itex]g \colon X \to Y[/itex] be functions continuous at [itex]x_0[/itex] such that [itex]f(x) = g(x)[/itex] for all [itex]x \ne x_0[/itex] in some neighborhood [itex]U[/itex] of [itex]x_0[/itex]. Then [itex]f(x_0) = g(x_0)[/itex].

Proof. Suppose [itex]f(x_0) \ne g(x_0)[/itex]. Then since [itex]Y[/itex] is Hausdorff, there exist disjoint open sets [itex]V_1, V_2 \subseteq Y[/itex] such that [itex]f(x_0) \in V_1[/itex] and [itex]g(x_0) \in V_2[/itex]. Since both [itex]f[/itex] and [itex]g[/itex] are continuous at [itex]x_0[/itex], there exist open sets [itex]W_1, W_2 \subseteq X[/itex] containing [itex]x_0[/itex] such that [itex]f(W_1) \subseteq V_1[/itex] and [itex]g(W_2) \subseteq V_2[/itex]. Let [itex]W = U \cap W_1 \cap W_2[/itex]; this set is open and it contains [itex]x_0[/itex]. Thus it contains some point [itex]x \ne x_0[/itex], since [itex]x_0[/itex] is a limit point of [itex]X[/itex]; we have [itex]f(x) = g(x)[/itex]. But [itex]f(W) \subseteq V_1[/itex] and [itex]g(W) \subseteq V_2[/itex], so [itex]f(W)[/itex] and [itex]g(W)[/itex] are disjoint, contradicting [itex]f(x) = g(x)[/itex].
 
  • #53
Vid
401
0
No the function doesn't have to be continuous.

Let f:R->R be any function. No constraints except that its a function. Tell me every value of f except at x_o. Now, we take the set of all functions from R->R and define an equivalence relation where f~g if f and g differ at finitely many points. Take any member g from the equivalence class of f. Since f and g differ on a set of measure zero, f(x_o) = g(x_o) with probability 1.

Of course the fact that we can choose g relies on the axiom of choice.
:D
 
  • #54
CRGreathouse
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Since f and g differ on a set of measure zero, f(x_o) = g(x_o) with probability 1.

I'd think that at a random point x, f(x) = g(x) with probability 1, but that wouldn't day anything about f(x_0) = g(x_0). Otherwise consider where f(x) = 1 for x = 0 and 0 otherwise, and g is uniformly 0. The probability that they agree on a random point is 1; the probability that they agree at x = 0 is 0.

This reminds me of a paper I read a few years ago, with a title along the lines of 'using the Axiom of Choice to see the future', which discussed similar techniques (as I recall!) to take a glance at a time epsilon in the future.
 
  • #55
Vid
401
0
Except we don't know anything about g except that it differs from f on a set of measure zero. Sure you could construct a counterexample, but it doesn't change the fact that if I have two functions that differ on a set of measure zero the probability that f(x_o) = g(x_o) is 1 for any x_o. Something happening with probability one isn't the same as something always happening.
 
  • #56
682
1
a counterintuitive piece of math... well, i read something about continuum hypothesis, and it says that it is proved that it cannot be proved or disproved based on the ZFC set theory. Isn't that weird?!

Also, any set can be well ordered by axiom of choice. And everything about the minimal uncountable (well-ordered) set.
 
  • #57
534
1
No the function doesn't have to be continuous.

Let f:R->R be any function. No constraints except that its a function. Tell me every value of f except at x_o. Now, we take the set of all functions from R->R and define an equivalence relation where f~g if f and g differ at finitely many points. Take any member g from the equivalence class of f. Since f and g differ on a set of measure zero, f(x_o) = g(x_o) with probability 1.

Of course the fact that we can choose g relies on the axiom of choice.
:D

Quite interesting. I'm a little sceptical about that, but I'll think about it some more.
 
  • #58
Hurkyl
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Since f and g differ on a set of measure zero, f(x_o) = g(x_o) with probability 1.
Why? This certainly doesn't follow from anything you said previously, because this is the first time you even mention probability. You never bother to specify what probability distribution you're using either.... (nor what the outcomes and events are)
 
  • #59
534
1
Alright, here's a specific example showing that the proof doesn't work.

Say you have two functions f and h that differ only at x0; then f ~ h. Choose any member g from [f] = [h]; by your argument, f(x0) = g(x0) with probability 1, and h(x0) = g(x0) with probability 1. But these are disjoint events, so that is absurd.
 
  • #60
CRGreathouse
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Alright, here's a specific example showing that the proof doesn't work.

Say you have two functions f and h that differ only at x0; then f ~ h. Choose any member g from [f] = [h]; by your argument, f(x0) = g(x0) with probability 1, and h(x0) = g(x0) with probability 1. But these are disjoint events, so that is absurd.

Heh, I said the same thing in post #54.
 
  • #61
534
1
No, not quite, since you're not picking a random function from [f]. I think what Vid was saying was that you should pick a random function from [f]. In any case, it doesn't work; that's what I proved in my post.
 
  • #62
90
5
I think few "counterintuitive" math problems remain "counterintuitive" when you at last understand them. Maybe "counterintuitive" before you have grasped it, but not after you have been aquainted to the concept.

But still there is one thing I find counterintitive more than anything I have learnt, although
belonging to physics rather than mathematics: Bernoullis fluid law, where static pressure
gets lower in more narrow passages while the speed of fluid at the same time is higher. I understand the theory, but still I find it very counterintuitive - not the least in practice, where you watch the flowing river surface sinking above for instance a bottom stone. :surprised
 
  • #63
Hurkyl
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There's a joke that says there exist only two kinds of theorems in mathematics: trivial ones, and deep ones. Trivial theorems are those I undersand, deep theorems are those I don't.
 
  • #64
CRGreathouse
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Let f:R->R be any function. No constraints except that its a function. Tell me every value of f except at x_o. Now, we take the set of all functions from R->R and define an equivalence relation where f~g if f and g differ at finitely many points. Take any member g from the equivalence class of f. Since f and g differ on a set of measure zero, f(x_o) = g(x_o) with probability 1.

Of course the fact that we can choose g relies on the axiom of choice.
:D

This reminds me of a paper I read a few years ago, with a title along the lines of 'using the Axiom of Choice to see the future', which discussed similar techniques (as I recall!) to take a glance at a time epsilon in the future.

I found the paper I was thinking of. Citation and quote:
Christopher Hardin and Alan D. Taylor, "A Peculiar Connection Between the Axiom of Choice and Predicting the Future" (2006):
Specifically, given the values of a function on an interval (−∞, t), the strategy produces a guess for the values of the function on [t,∞), and at all but countably many t, there is an ε > 0 such that the prediction is valid on [t, t + ε). Noting that any countable set of reals has measure 0, we can restate this informally: at almost every instant t, the strategy predicts some “ε-glimpse” of the future.
 
  • #66
HallsofIvy
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Of course, the word "converge" is (intentionally) incorrect there.
 
  • #67
2
0
How about the sum of all positive integers s= 1 + 2 + 3 + 4 + ... = -1/12
 
  • #68
Char. Limit
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How about the sum of all positive integers s= 1 + 2 + 3 + 4 + ... = -1/12

Technically that's not true, at least not in terms of a normal sum. I believe it is true if you consider the Abel sum of all positive integers.
 
  • #69
2
0
Technically that's not true, at least not in terms of a normal sum. I believe it is true if you consider the Abel sum of all positive integers.

It's not Abel summable, but there are several ways to derive the result such as analytic continuation of the Riemann zeta function or Ramanujan Summation, but the simplest was proposed by Euler. Perhaps more interestingly, it has been experimentally measured, in the Casimir effect for example.
 

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