# Thread: Continuity mk2

1. ## Continuity mk2

Prove that $\displaystyle f(x) = \begin{cases} x \ \ \ \ \text{if} \ x \ \ \text{is irrational} \\ 0 \ \ \ \ \text{if} \ x \ \text{is rational} \end{cases}$ is discontinuous at $\displaystyle x_0 \neq 0$.
I need to find $\displaystyle |f(x)-f(x_0)|< \epsilon$ when $\displaystyle 0<|x-x_0|< \delta$ (where $\displaystyle \delta,\epsilon>0$).

Suppose that $\displaystyle x_0$ is rational and $\displaystyle x$ is rational. Therefore:

$\displaystyle |f(x)-f(x_0)|< \epsilon \Leftrightarrow \ |f(x)|< \epsilon \Leftrightarrow \ 0< \epsilon$.
This is valid from the definition of $\displaystyle \epsilon$.

Suppose that $\displaystyle x_0$ is rational and $\displaystyle x$ is irrational.

$\displaystyle |f(x)-f(x_0)|< \epsilon \Leftrightarrow \ |f(x)|<\epsilon \Leftrightarrow \ x< \delta$.
Hence take $\displaystyle \delta=\epsilon$.

So the proof for when $\displaystyle x_0$ is rational is as follows:

Let $\displaystyle \epsilon>0$, $\displaystyle x_0$ be rational and $\displaystyle \delta:=\epsilon$. When $\displaystyle 0<|x-x_0|< \delta$ I have:

$\displaystyle |f(x)-f(x_0)|=|f(x)|< \epsilon$ as required.

Suppose that $\displaystyle x_0$ is irrational and $\displaystyle x$ is also irrational.

$\displaystyle |f(x)-f(x_0)|<\epsilon \Leftrightarrow \ |f(x)-x_0|< \epsilon \Leftrightarrow \ |x-x_0|<\epsilon$.
Therefore pick $\displaystyle \delta=\epsilon$ as before.

Suppose that $\displaystyle x_0$ is irrational and $\displaystyle x$ is rational.

$\displaystyle |f(x)-f(x_0)|<\epsilon \Leftrightarrow \ |f(x)-x_0|< \epsilon \Leftrightarrow |-x_0|<\epsilon \Leftrightarrow \ |x_0|<\epsilon$.

At this point, a $\displaystyle \delta$ value cannot be chosen since it isn't in the above inequality chain. Since $\displaystyle |x_0|<\epsilon$ has to be valid $\displaystyle \forall \epsilon>0$, there is a clear contradiction (eg, if $\displaystyle x_0=5$ I can make $\displaystyle \epsilon=4$).

Is this sufficient proof to show that $\displaystyle f(x)$ is discontinuous at $\displaystyle x_0 \neq 0$?

(btw, I have shown that $\displaystyle f(x)$ is continuous at $\displaystyle x_0=0).$

PS. Can someone show me a way of doing this with sequences? This $\displaystyle \epsilon, \delta$ method is a little long.

2. Theorem: If $\displaystyle f$ is continuous at $\displaystyle a$ and $\displaystyle \left| {f(a)} \right| > 0\, \Rightarrow \,\left( {\exists \delta } \right)\left[ {x \in \left( {a - \delta ,a + \delta } \right)\, \Rightarrow \,\left| {f(x)} \right| > 0} \right]$.
In other words, if a continuous at a point and is not zero there then there is an open interval about the point on which the function is not zero.

Applying that theorem to this problem quickly gives the result.
Note in any open interval there are irrational numbers and at irrational numbers the function is zero.
That contradicts the theorem.

3. hmm, I think I need a little help with this one:

So I need to use $\displaystyle |f(a)|>\epsilon$ when $\displaystyle |x-a|<\delta$ to achieve a contradiction of some sort.

Let a be irrational.

$\displaystyle |f(a)|>\epsilon \Leftrightarrow|a|>\epsilon \Leftrightarrow \ -\epsilon<a<\epsilon \Leftrightarrow -\epsilon<-a<\epsilon \Leftrightarrow x-\epsilon<x-a<x+\epsilon$.

What's the next step from here? Comparing with $\displaystyle |x-a|<\delta$ seems to suggest $\displaystyle \delta=x+\epsilon$, which is clearly wrong, but at the same time it isn't a contradiction.

4. Why can't you use open balls to prove this? Consider, in the image of f, the set A={f(x)| 1<=f(x)<=2 }. Because neither 1, nor 2, are irrational or 0, they are not members of the image of f, so they are not members of A. So A is an open set. But the preimage of A is going to be [1,2], which is closed. But f is continuous iff every preimage of an open set is open itself. Is that right?

5. umm, you probably can.

I've never learnt about open balls (to wiki!!), only about epsilon-delta proofs and sequences. For this question, continuity at 0 is proved using epsilon-delta so i'm trying to prove discontinuity at other points using this method.

6. Originally Posted by Showcase_22
only about epsilon-delta proofs and sequences. For this question, continuity at 0 is proved using epsilon-delta so i'm trying to prove discontinuity at other points using this method.
Use the given function $\displaystyle f(x) = \left\{ {\begin{array}{ll} {x,} & {x \text{ is rational}} \\ {0,} & { x \text{ is irrational}} \\ \end{array} } \right.$.
Suppose that this function continuous at $\displaystyle a\ne 0$ and $\displaystyle \left| {f(a)} \right| > 0$.
That means that $\displaystyle a$ is rational, why?
Using the definition of continuity, $\displaystyle \left( {\exists \delta > 0} \right)\left[ {\left| {x - a} \right| < \delta \, \Rightarrow \,\left| {f(x) - f(a)} \right| < \frac{{\left| {f(a)} \right|}}{2}} \right]$
It follows that $\displaystyle \left| {\left| {f(x)} \right| - \left| {f(a)} \right|} \right| \leqslant \,\left| {f(x) - f(a)} \right| < \frac{{\left| {f(a)} \right|}}{2}\, \Rightarrow \,\left| {f(x)} \right| > \,\frac{{\left| {f(a)} \right|}}{2} > 0$.
In other words, $\displaystyle \left( {\forall x \in \left( {a - \delta ,a + \delta } \right)} \right)\left[ {|f(x)| > 0} \right]$.
That is impossible because between any two numbers there is an irrational number at which $\displaystyle f$ must be zero.

On the other hand, suppose the function continuous at $\displaystyle c \ne 0$ and $\displaystyle \left| {f(c)} \right| = 0$.
That means that $\displaystyle c$ is irrational, why? We may assume that $\displaystyle c > 0$.
Now $\displaystyle 0 < \frac{c}{2} < c\;\& \,\left( {\exists r \in \mathbb{Q}} \right)\left[ {\frac{c}{2} < r < c} \right]$.
Using $\displaystyle \varepsilon = \frac{c}{2}$ it is impossible to find a $\displaystyle \delta$ that works.

P.S. In any $\displaystyle \varepsilon -$neighborhood $\displaystyle \left( {c - \varepsilon ,c + \varepsilon } \right)$ of $\displaystyle c$, $\displaystyle \left( {\exists q \in \mathbb{Q}} \right)\left[ {c < q < c + \varepsilon } \right]$.
That implies that $\displaystyle c < q = f(q)$.
That means that there is no $\displaystyle \delta$ that works using the definition if we take $\displaystyle {\varepsilon = c > 0}$. (if $\displaystyle c<0$ a similar trick still works).

7. $\displaystyle \left( {\exists \delta > 0} \right)\left[ {\left| {x - a} \right| < \delta \, \Rightarrow \,\left| {f(x) - f(a)} \right| < \frac{{\left| {f(a)} \right|}}{2}} \right]$?
How did you get the above?

$\displaystyle \left| {\left| {f(x)} \right| - \left| {f(a)} \right|} \right| \leqslant \,\left| {f(x) - f(a)} \right| < \frac{{\left| {f(a)} \right|}}{2}\, \Rightarrow \,\left| {f(x)} \right| > \,\frac{{\left| {f(a)} \right|}}{2} > 0$
I also can't see how one side implies the other. Did you minus something from both sides?

It is impossible to find a $\displaystyle \delta$ that works.
Why is that the case?

(sorry for the barrage of questions!)

8. Sorry to double post but I needed the toolbar at the top of the text box.

I think i'm starting to get this, so i'll have a go at answering my own question (thanks for answering one!):

$\displaystyle \left| {\left| {f(x)} \right| - \left| {f(a)} \right|} \right| \leqslant \,\left| {f(x) - f(a)} \right| < \frac{{\left| {f(a)} \right|}}{2}\, \Rightarrow \,\left| {f(x)} \right| > \,\frac{{\left| {f(a)} \right|}}{2} > 0$

I can't see how one side implies the other.
x can either be rational or irrational.
If x is rational we get $\displaystyle |x-a|< \delta$.
If x is irrational we get $\displaystyle |0-a|=|a|=|f(a)|$.

From the triangle inequality we know that $\displaystyle ||f(x)|-|f(a)|| \leq |f(x)-f(a)| \leq |f(a)|$ (the last step was attained from above).

$\displaystyle \Rightarrow 0< |f(x)|-|f(a)|<||f(x)|-|f(a)|| \leq |f(x)-f(a)| \leq |f(a)|$

$\displaystyle \Rightarrow 0< |f(x)|-|f(a)| \leq |f(a)|$

$\displaystyle \Rightarrow 0<|f(a)|<|f(x)| \leq 2|f(a)|$ (adding f(a) to all the sides and making it greater than 0).

but $\displaystyle 0<\frac{|f(a)|}{2}<|f(a)|$ so putting this in the inequality gives:

$\displaystyle \Rightarrow 0<\frac{|f(a)|}{2}<|f(a)|<|f(x)| \leq 2|f(a)|$

and isolating the bits we need gives the required result $\displaystyle 0<\frac{|f(a)|}{2}<|f(x)|$