## 2012-05-28

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.

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