Continuity mk2

**Showcase_22** · March 18th 2009, 12:08 PM

Prove that $f(x) = \begin{cases} x \ \ \ \ \text{if} \ x \ \ \text{is irrational} \\ 0 \ \ \ \ \text{if} \ x \ \text{is rational} \end{cases}$ is discontinuous at $x_0 \neq 0$ .

I need to find $|f(x)-f(x_0)|< \epsilon$ when $0<|x-x_0|< \delta$ (where $\delta,\epsilon>0$ ).

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

**Plato** · March 18th 2009, 12:28 PM

Theorem: If $f$ is continuous at $a$ and $\left| {f(a)} \right| > 0\, \Rightarrow \,\left( {\exists \delta } \right)\left[ {x \in \left( {a - \delta ,a + \delta } \right)\, \Rightarrow \,\left| {f(x)} \right| > 0} \right]$ .
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.

**Showcase_22** · March 19th 2009, 02:34 AM

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

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

Let a be irrational.

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

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

**mr. frege** · March 19th 2009, 05:35 AM

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?

**Showcase_22** · March 19th 2009, 06:44 AM

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.

**Plato** · March 19th 2009, 08:02 AM

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 $f(x) = \left\{ {\begin{array}{ll} {x,} & {x \text{ is rational}} \\ {0,} & { x \text{ is irrational}} \\ \end{array} } \right.$ .
Suppose that this function continuous at $a\ne 0$ and $\left| {f(a)} \right| > 0$ .
That means that $a$ is rational, why?
Using the definition of continuity, $\left( {\exists \delta > 0} \right)\left[ {\left| {x - a} \right| < \delta \, \Rightarrow \,\left| {f(x) - f(a)} \right| < \frac{{\left| {f(a)} \right|}}{2}} \right]$
It follows that $\left| {\left| {f(x)} \right| - \left| {f(a)} \right|} \right| \leqslant \,\left| {f(x) - f(a)} \right| < \frac{{\left| {f(a)} \right|}}{2}\, \Rightarrow \,\left| {f(x)} \right| > \,\frac{{\left| {f(a)} \right|}}{2} > 0$ .
In other words, $\left( {\forall x \in \left( {a - \delta ,a + \delta } \right)} \right)\left[ {|f(x)| > 0} \right]$ .
That is impossible because between any two numbers there is an irrational number at which $f$ must be zero.

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

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

**Showcase_22** · March 19th 2009, 09:33 AM

$ \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?

$ \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 $\delta$ that works.

Why is that the case?

(sorry for the barrage of questions!)

**Showcase_22** · March 21st 2009, 04:52 AM

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!):

$ \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 $|x-a|< \delta$ .
If x is irrational we get $|0-a|=|a|=|f(a)|$ .

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

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

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

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

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

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

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

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