Question

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

I will provide a hint, and later a solution if no one can find one.

Answer 2

\[f(x)=x, [-1,-1] \]
lol

Answer 3

Sorry, I don't get it

Answer 4

That's not in the interval he gave.

Answer 5

oh oops

Answer 6

Let me know when you'd like a hint (if you'd like one).

Answer 7

Think of what you can do with \(\mathbb{Q}\)

Answer 8

How about you start by thinking of a function from that interval to R, that is continuous nowhere.

Answer 9

should i give it away

Answer 10

No

Answer 11

I'm going to give a hint by defining a function continuous nowhere. One second...

Answer 12

\[
f(x) = \left\{
\begin{array}{c|c}
1 & x \in \mathbb{Q}\\
0 & x \notin \mathbb{Q}
\end{array}\right.
\]

Answer 13

\[f(x)=\frac{C}{x-x}\]This one is messed up everywhere.

Answer 14

That's not defined anywhere. The problem states that the function must be defined on the interval.

Answer 15

that is not defined on [0,1]

Answer 16

I know it's not... I'm just brainstorming. xd

Answer 17

Start by having a look at my function above. You can play around with it to get the solution.

Answer 18

modify alchemista's hint

Answer 19

What's the syntax for those brackets?

Answer 20

The source for my function:
f(x) = \left\{
\begin{array }{c|c}
1 & x \in \mathbb{Q}\\
0 & x \notin \mathbb{Q}
\end{array}\right.

Answer 21

Remove the space in \begin{array }

Answer 22

You can right click the source for anyone's latex to show the source and copy it fyi

Answer 23

sec... syntax errors...

Answer 24

If you've thought of a solution, just paste the source and I'll correct the errors in the LaTeX

Answer 25

When you are thinking about this, remember about the delta-epsilon definition of continuity.

\(f(x)\) is continuous at \(a\) if for every \(\varepsilon >0\) there is a \(\delta > 0\) such that when \(|x - a| < \delta\) then \(|f(x) - f(a)| < \varepsilon\)

Answer 26

\[ f(x) =\left \{ \begin {array} {c|c} 1 & x \in [0.5, 0.5] \\ 0 & x \notin (-\infty, 0) \cup (1, \infty) \end {array} \right \]

Answer 27

ich gebe auf xd

Answer 28

\[f(x) =\left \{ \begin {array} {c|c}
1 & x \in [0.5, 0.5] \\
0 & x \notin (-\infty, 0) \cup (1, \infty)
\end {array} \right.\]

Answer 29

This?

Answer 30

The problem with your function is, it is only defined at 0.5 on the required interval.

Answer 31

The function must be defined everywhere on [0,1] but must be continuous only at one point.

Answer 32

http://openstudy.com/groups/mathematics/updates/4e45fb760b8b3609c724a595#/groups/mathematics/updates/4e460a110b8b3609c724bc7a