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mathematics community!

I'm teaching a course in Real Analysis soon, and one thing I wanted to include were a few "fake proofs" for my students to evaluate. The research I've done hasn't turned up any fake-proofs regarding continuity.

So I'll put it out to you: do any of you have a favorite "proof" involving continuity or a sequential criterion with a flaw in it that could potentially stump some people?

Steve
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    "All horses are the same color" is a classic. – Doug M Aug 14 '18 at 20:01
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    you may find some good examples here – lulu Aug 14 '18 at 20:03
  • Limit of a pointwise convergent sequence of continuous functions is continuous. – Somos Aug 14 '18 at 21:36
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    @DougM : In 1997 I took a real analysis class and one of the homework problems was to evaluate the fake induction proof of the statement "All horses are brown." I had no idea it was a famous problem until your comment above. I am amazed that in 21 years the problem has not changed from "horses" to something else, like Pokemon characters! – Michael Aug 15 '18 at 04:57
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    I never heard it as horses until about five years ago. It was always marbles in a bag. And that was 38 years ago. – fleablood Aug 15 '18 at 05:22
  • @fleablood : It sounds like you have lost your marbles. – Michael Aug 15 '18 at 05:26
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    Well, if the proof works on my marbles, then I didn't have very many to begin with. – fleablood Aug 15 '18 at 05:30

2 Answers2

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This does not involve continuity, but it is a "false proof" that involves basic concepts. I made this up as a problem set question when I first taught linear algebra.

What is incorrect about the following chain of reasoning?

We have this problem: $$ \left[ \begin{array}{cc} 1 & -2 \\ 3 & 2 \\ 1 & -4 \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{c} -1 \\ 5 \\ 1 \end{array} \right] $$

Hence: $$ \left[ \begin{array}{ccc} 2 & 6 & 2 \\ -3 & 1 & 0 \end{array} \right] \left[ \begin{array}{cc} 1 & -2 \\ 3 & 2 \\ 1 & -4 \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{ccc} 2 & 6 & 2 \\ -3 & 1 & 0 \end{array} \right] \left[ \begin{array}{c} -1 \\ 5 \\ 1 \end{array} \right] $$

Hence: $$ \left[ \begin{array}{cc} 22 & 0 \\ 0 & 8 \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array}\right] = \left[ \begin{array}{c} 30 \\ 8 \end{array} \right] $$

Therefore, the solution is $y = 1$, $x = 30/22$.

Michael
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  • Because the initial system had a contradiction, right ? – Our Aug 15 '18 at 05:20
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    @onurcanbektas : Yep. It is actually a correct proof of the (true) statement “If $(x,y)$ satisfies that first equation, then $(x,y)$ must be $(30/22, 1)$.” Of course, there is no $(x,y)$ that satisfies that first equation, so the “then” part is irrelevant! – Michael Aug 15 '18 at 05:23
  • Thank for the response; It is indeed a nice "false proof". – Our Aug 15 '18 at 05:25
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I'm not sure I have a favorite one, but I find this one nice:

The proof of the following statement has a flaw. Identify the false statement in the proof, and give an example to show that it is false:

Every bounded continuous real-valued function $f$ on $\mathbb{R}$ attains its maximum.

Proof. Let $M=\sup\{f(x) \colon x \in \mathbb{R}\}$, and let $x^*, x_n \in \mathbb{R}$ such that $x_n \to x^*$ and $f(x_n) \to M$. Since $f$ is continuous, $f(x_n) \to f(x^*)$, which implies $f(x^*)=M$. Hence, $x^*$ is where $f$ attains its maximum.

In general, a nice place for ideas or even outright problems might be Counterexamples in Calculus.

mathematics2x2life
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    Nice one. This is almost like "proof by assuming the conclusion." – Michael Aug 15 '18 at 05:15
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    I´m not sure if I got it, but the only step that looks ilegal to me is assuming there is a sequence such that $x_n \to x^*$ and $f(x_n) \to M$. My counterexample would be $\arctan$... –  Aug 15 '18 at 06:58
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    @dude3221 The error lies precisely there. Essentially, you've prevented $x_n$ from converging to $x^$ by 'pushing' $x^$ out to $\infty$. Your example of $\arctan \theta$ works just fine, as would any function (meeting the theorem conditions) whose image has a horizontal asymptote. – mathematics2x2life Aug 16 '18 at 02:57