The following function fails the vertical line test, lol

\displaystyle f(x) = \begin{cases} 0 & \text{if } x \in \mathbb{Q} \\ 1 & \text{if } x \in \mathbb{R} \setminus \mathbb{Q}. \end{cases}

OK, maybe that’s up for philosophical debate…

EDIT: \forall PMATH 352 / 740 students: the lectures for Monday and Wednesday of this week are in MC 5136B.

\forall MATH 245 students: the lecture for Monday is in AL 201. The lecture for Wednesday is in RCH 208, as far as I know.

\forall CO 342 students: the lecture for Wednesday is in MC 4040.

\forall CS 341 students: the lecture for Tuesday is in RCH 309.


About mlbaker

just another guy trying to make the diagrams commute.
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3 Responses to 2012-05-28

  1. Joseph says:

    PMath 352 is there on Wednesday, too.

  2. Anon. says:

    How does f(x) fail the test. I don’t see 2 points
    “crossing” the line.

    • mlbaker says:

      you’re thinking too much like a mathematician; think like a high school student with a pencil and paper

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