Question

Continuity problem continued (old thread getting slow):

Define \(f:[0,1] \to \mathbb{R}\) such that \(f\) is only continuous at one point. Furthermore, prove that the function you have provided is only continuous at that single point.

Answer 1

If you'd like I can give it away.

Answer 2

bump..(so i get notified if someone posts)

Answer 3

Think about having a set of point discontinuities converging to your desired point. You want to satisfy the requirement where for any epsilon neighbourhood around the image of the point you can find a neighbourhood around the point such that image of that neighbourhood is contained in the epsilon neighbourhood.

Answer 4

Should I paste my solution or shall I wait?

Answer 5

\[0=x_{1},x_{2},...,x_{n-1},x_{n}=1\]\[n\rightarrow\infty\]\[f(x)=\left\{\begin{matrix}1-2x, & x\in \left\{ x_{2i} \right\} \\ 2x-1, & x\notin \left\{ x_{2i} \right\}\end{matrix}\right.\]

Answer 6

You have the right idea, but the sequence is a bit problematic.

Answer 7

\[f(x)=\left\{\begin{matrix}1-2x, & x\in \left\{x_{2i}|1 \le i \le n\right\} \\ 2x-1, & x\notin \left\{ x_{2i}|1 \le i \le n\right\}\end{matrix}\right.\]

Answer 8

yeah xd

Answer 9

The problem is you need a set of points that is dense on the reals.

Answer 10

That's why you should be using \(\mathbb{Q}\) (the rationals).

Answer 11

Here I will post my full solution.

Answer 12

An example of a function satisfying the requirements is as follows:
\[f(x) = \left\{
\begin{array}{c|c}
x & x \in \mathbb{Q}\\
0 & x \notin \mathbb{Q}
\end{array}\right.\]
The function is continuous at \(0\) but discontinuous everywhere else.

We want to show that for every \(\varepsilon > 0\) there is a \(\delta > 0\) such that when \(|x| < \delta\) then \(|f(x)| < \varepsilon\).

Simply take \(\delta = \epsilon\). Since \(f(x)\) is 0 or \(x\) when \(|x| < \varepsilon\) then \(|f(x)| < \varepsilon\)

We also want to show that for every \(a \in (0, 1]\) the function is discontinuous. So we want to show that there is an \(\varepsilon > 0\) such that for every \(\delta > 0\) there exists \(c \in (a -\delta, a + \delta)\) such that \(|f(c) - f(a)| \geq \varepsilon\).

Just take \(0 < \epsilon < \frac{a}{2}\), I am going to hand wave and say this works since if you take a point in \(\mathbb{Q}\) the epsilon neighbourhood won't include the irrational points at zero, and if you take an irrational point the neighbourhood won't include the rational points on the line \(x\).

If you want me to write that out rigorously let me know.

Answer 13

You were on the right track.

Answer 14

i don't get it

Answer 15

The function is defined such that the irrationals are 0 and the rationals are on the line \(y=x\)

Answer 16

You have a set of point discontinuities converging to 0.

Answer 17

I enjoy this type of exercises. You should keep a daily or something.
I'm off to bed -_- goodnight!

Answer 18

ok

Answer 19

Do you see how you can satisfy the negation of the continuity conditions everywhere but at 0?

Answer 20

yes
i definitely would not have thought the function satisfying the conditions you ask to satisfy