Question

In need of a quick hw check: Find an ex. of a discontinuous fn s.t. \(f:[0,1]\rightarrow R\) where the Intermediate Value Theorem fails.

Answer 1

So I defined \[f(x)=[0 ~if~ x\in Q ~and~ 0\le x\le 1, ~~1~if~x\not{\in} Q ~and~0\le x\le 1]\]

then for any y s.t. 0<y<1 the IVT fails.
Does this work? Is it actually answering the question?

Answer 2

@jim_thompson5910 , have you had analysis?

Answer 3

the first function i think of is 1/(1-2x)

Answer 4

why 2x?

Answer 5

this function has no zeroes eventhough it changes sign in (0,1)

Answer 6

i was just thinking my answer doesn't fit the problem and was at 1/(.5-x)

Answer 7

i think we want discontinuity somewhere between (0,1) right

Answer 8

well, it could be discontinuous anywhere

Answer 9

but the issue is thinking of a fn that maps 0,1 onto all of the reals

Answer 10

Oh you want an onto function is it ?

Answer 11

yea, my function doesn't satisfy the conditions I don't think

Answer 12

your function is fine for the conditions stated, but it is NOT surjective.

Answer 13

well, I mean the conditions make it have to be surjective, don't they?

Answer 14

why? \(f:A\rightarrow B\) does not imply \(f\) is onto \(B\) it just implies \(B\) is the codomain.

Answer 15

You do not read that as [0,1] is mapped onto the reals?

Answer 16

nope

Answer 17

oh, that's how I read it

Answer 18

http://en.wikipedia.org/wiki/Function_(mathematics)

scroll down to notation.

Sometimes people use a double arrow to indicate onto, but there is no standard, because we use that same double arrow to mean "the trivial map".

Answer 19

...hmm I have always read it as onto. Maybe not spanning now that I think of it though

Answer 20

alright, so I can use this. Thanks for the clarification. I have another which as a technicality is bothering me, can I tag you?

Answer 21

Unless your book/teacher uses that notation to be onto, but I have never seen that. Notice also that @ganeshie8 also did not assume surjectivity.

Answer 22

true ok I guess I was just so used to reading it as onto for some reason

Answer 23

it would make more sense... more bad notation in mathematics.

Answer 24

true, well hopefully it works because I just can't think of a disc. fn with that domain which could span the reals...unless it was like a point disc