Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How can I prove that the function $f:[0,1] \rightarrow \mathbb{R}$, defined as $$ f(x) = \left\{\begin{array}{l l} x &\text{if }x \in \mathbb{Q} \\ x^2 & \text{if } x \notin \mathbb{Q} \end{array} \right. $$

is continuous on $0$ and $1$, but nowhere else?

I really don't know where to start.

I know the (equivalent) definitions of continuous functions (epsilon delta, $\lim f(x) = f(c)$, topological definition with epsilon and delta neighbourhoods, and the definition where as $x_n$ goes to $c$, it implies that $f(x_n)$ goes to $f(c)$).

share|improve this question

2 Answers 2

up vote 7 down vote accepted

Hint: Choose a sequence of rationals converging to an irrational and vice-versa and recall
that continuity also implies sequential continuity, to conclude what you want.

Move your mouse over the gray area for a complete solution.

Consider $a \in [0,1] \backslash \mathbb{Q}$. For this $a$, choose a sequence of rationals converging to $a$ i.e. $\{a_n\}_{n=1}^{\infty}$, where $a_n \in [0,1] \cap\mathbb{Q}$. One such choice for this sequence is $a_n = \dfrac{\lfloor 10^n a\rfloor}{10^n}$. If $f$ were to be continuous (recall that continuity also implies sequential continuity), then $$\lim_{n \to \infty} f(a_n) = f\left(\lim_{n \to \infty} a_n \right) = f(a)$$ But this gives us that $a = a^2$, which is not true for any $a \in [0,1] \backslash \mathbb{Q}$. Similarly, argue when $a \in [0,1] \cap\mathbb{Q}$, by picking a sequence of irrationals converging to $a$ i.e. $\{a_n\}_{n=1}^{\infty}$, where $a_n \in [0,1] \backslash \mathbb{Q}$. One such choice is $a_n = \left(1 - \dfrac{\sqrt{2}}{2n} \right)a$. If $f$ were to be continuous (recall that continuity also implies sequential continuity), then $$\lim_{n \to \infty} f(a_n) = f \left(\lim_{n \to \infty} a_n \right) = f(a)$$ But this gives us that $a^2 = a$, which is true only for any $a =0$ or $a=1$. Hence, the function is continuous only at $0$ and $1$.

share|improve this answer
    
Thank you very much! I do not need to define a specific sequence $a_n$ right? Really appreciate your help, thanks!! –  MSKfdaswplwq Oct 20 '12 at 20:42
1  
@Hempo You do not need to define a specific sequence. You just need the fact that any number can be approached either as sequence of purely rational numbers or purely irrational numbers. –  user17762 Oct 20 '12 at 20:47

Well I guess it should be something to do with the fact that $x = x^2$ has solutions $0$ and $1$.

Let $a>1$ be rational, then we can always find an irrational $a' > a$ very close to $a$ such that $|f(a)-f(a')| = |a-a'^2| > a(a-1)$.

You can do the same when $a$ is irrational, then you can do it when $a<0$ and also $0 < a < 1$.

Now we just want to prove continuous at $0$ and $1$. Ill show how for $1$:

We need to show the definition of continuity: $\forall \epsilon > 0,\,\, \exists \delta,\,\, \forall x,\,\, |x-1| < \delta \to |f(x) - f(1)| < \epsilon$

Note by cases that if $x < 1$ or $x > 1$ then $|x^2-1| > |x-1|$ so we can prove continuity by proving the stronger statementn $\forall \epsilon > 0,\,\, \exists \delta,\,\, \forall x,\,\, |x-1| < \delta \to |x^2 - 1| < \epsilon$ and this isn't difficult, just factor $|x^2 - 1^2| = |x-1||x+1|$ the first term is $\delta$ and the second is $2$, so (at least for $x$ relatively close to $1$, which is all that matters) use $\varepsilon = \delta/3$ and the proof goes through.

By the way, $x^2$ repels around 1 but it attracts around 0 so for the proof in the 0 case you can just use $\varepsilon = \delta$.

share|improve this answer
1  
$$\vert a - a'^2 \vert \leq \vert a - a^2 \vert + \vert a^2 - a'^2 \vert$$ –  user17762 Oct 20 '12 at 20:48
    
@Marvis, thanks I corrected it. –  sperners lemma Oct 20 '12 at 21:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.