continuity for a real-valued function

**thedoctor818** · February 22nd 2010, 11:27 AM

I have one more question for now at least. My task is as follows.

Let $f \in \mathbb{R} \wedge \text{be continuous } \backepsilon dom f = (a,b).$ Show that if $f(r) = 0 \, \text{for each } r \in \mathbb{Q} \backepsilon r \in (a,b), \text{then } f(x) = 0 \forall x \in (a,b).$

**maddas** · February 22nd 2010, 01:21 PM

Use that the rationals are dense in $\mathbb{R}$ by chusing for every x in the domain, a sequence of rational numbers $x_n$ in the domain converging to x. Since f is continuous, $\lim_{n\to \infty} f(x_n) = f(\lim_{n\to \infty}x_n)$ , and since the LHS is zero, the RHS is too.

**thedoctor818** · February 22nd 2010, 02:43 PM

I have heard of the topic of `denseness', but we have not formally discussed this concept in my analysis class. Is there another way of framing your reply, or perhaps could you restate the reply while including the definition of denseness???

Thanks,
-the Doctor

**Plato** · February 22nd 2010, 02:58 PM

Originally Posted by thedoctor818

I have heard of the topic of `denseness', but we have not formally discussed this concept in my analysis class. Is there another way of framing your reply, or perhaps could you restate the reply while including the definition of denseness???

In order to do this particular problem your class must have proven that:
between any two real numbers there is a rational number and an irrational number.
Have you proved that?

**thedoctor818** · February 22nd 2010, 04:16 PM

We have talked about this sort of informally, but not very formally or in an `official' manner.

**Plato** · February 22nd 2010, 04:38 PM

Originally Posted by thedoctor818

We have talked about this sort of informally, but not very formally or in an `official' manner.

In that case, You must prove the following two therorems.
Between any two real numbers there is a rational number.
Between any two real numbers there is an irrational number..

**thedoctor818** · February 22nd 2010, 06:30 PM

Please help. How do I prove or begin to prove those aforementioned theorems.

**Drexel28** · February 22nd 2010, 06:32 PM

Originally Posted by thedoctor818

Please help. How do I prove or begin to prove those aforementioned theorems.

Try this

**davismj** · February 22nd 2010, 06:45 PM

Originally Posted by thedoctor818

Please help. How do I prove or begin to prove those aforementioned theorems.

Have you covered the Archimedean Property? It states that

(a) For every real number x, there exists a natural number n such that n > x.
(b) Similarly, for every real number y, there exists a natural number m such that 1/m < y.

(a) Should be trivial. (b) Follows from (a) (raise both sides of the inequality to the (-1), and of course reverse the inequality).

**thedoctor818** · February 22nd 2010, 06:55 PM

So, if I have this correctly (in order to prove there exists a rational number between x & y):

Let $x,y \in \mathbb{R} \backepsilon x < y$ and consider $(x,y).$ We must find a rational number inside this interval. By the Archimedean Theorem, $\exists n \in \mathbb{N} \backepsilon n > \dfrac{1}{y-x}. \Rightarrow ny > nx + 1.$ We then find an $m \in \mathbb{Z} \backepsilon m \leq nx + 1 < m + 1.$ Now, we have that $m-1 \leq nx < ny. \Rightarrow x < \dfrac{m}{n} \leq x + \dfrac{1}{n} < y,$ thus proving an $\dfrac{m}{n} \in \mathbb{Q} \backepsilon \dfrac{m}{n} \in (x,y).$

Is that correct? If so, I am uncertain of how to prove this for $a \in \overline{\mathbb{Q}}.$

**davismj** · February 22nd 2010, 07:05 PM

Originally Posted by thedoctor818

So, if I have this correctly (in order to prove there exists a rational number between x & y):

Let $x,y \in \mathbb{R} \backepsilon x < y$ and consider $(x,y).$ We must find a rational number inside this interval. By the Archimedean Theorem, $\exists n \in \mathbb{N} \backepsilon n > \dfrac{1}{y-x}. \Rightarrow ny > nx + 1.$ We then find an $m \in \mathbb{Z} \backepsilon m \leq nx + 1 < m + 1.$ Now, we have that $m-1 \leq nx < ny. \Rightarrow x < \dfrac{m}{n} \leq x + \dfrac{1}{n} < y,$ thus proving an $\dfrac{m}{n} \in \mathbb{Q} \backepsilon \dfrac{m}{n} \in (x,y).$

Is that correct? If so, I am uncertain of how to prove this for $a \in \overline{\mathbb{Q}}.$

How about this. To prove that the Rationals (Q) are dense in the Reals (R), we must show that for two real numbers x and y, there exists a rational number q = m/n satisfying x < m/n < y.

By (a) we know that an m exists such that m > nx, which implies that x < m/n, the left half of the inequality. By (b) we know that an n exists such that 1/n < y/m, which implies that m/n < y, the right half of the inequality. Thus, there exists a rational number q = m/n x < m/n < y for any two arbitrary real numbers x and y.

For the irrationals, choose an irrational number w, and add w to x, m/n, y. Then, prove that m/n + w is no longer rational.

**thedoctor818** · February 22nd 2010, 07:10 PM

Thanks, I'll try that first thing in the morning. Once I have done that, however; I am not quite certain of the relationship of these 2 proofs to my problem though???

**davismj** · February 22nd 2010, 07:23 PM

Originally Posted by thedoctor818

Thanks, I'll try that first thing in the morning. Once I have done that, however; I am not quite certain of the relationship of these 2 proofs to my problem though???

Never seen that kind of question before (I'm in my undergraduate Real Analysis class at the moment). From what I can tell, it follows from the fact that for any irrational number y, there are two rational numbers p and q satisfying p < y < q. Since f(p) = f(q) = 0 and f(x) is continous on (a,b), f(y) must be 0 as well.

As for a formal proof, sorry.

**thedoctor818** · February 22nd 2010, 07:29 PM

I am also in my undergrad analysis class, & this is one of our problems. Thanks again for your help, & for the last bit - your intuition - maybe we can rework that into a formal proof. Thanks.

**Drexel28** · February 22nd 2010, 07:54 PM

Originally Posted by davismj

Never seen that kind of question before (I'm in my undergraduate Real Analysis class at the moment). From what I can tell, it follows from the fact that for any irrational number y, there are two rational numbers p and q satisfying p < y < q. Since f(p) = f(q) = 0 and f(x) is continous on (a,b), f(y) must be 0 as well.

As for a formal proof, sorry.

Originally Posted by thedoctor818

I am also in my undergrad analysis class, & this is one of our problems. Thanks again for your help, & for the last bit - your intuition - maybe we can rework that into a formal proof. Thanks.

What it means for $\mathbb{Q}$ to be dense in $\mathbb{R}$ is that ever point of $\mathbb{R}$ is a limit point of $\mathbb{Q}$ . Consequently, it easy then to extract a rational sequence $\{q_n\}$ such that $q_n\to x$ for ever $x\in \mathbb{R}$ . So, since your function is continuous we have that $x_n \to x\implies f(x_n)\to f(x)$ and so let $x\in (a,b)$ then there exists some $q_n\to x$ and so $f(x)=\lim\text{ }f(q_n)=\lim\text{ }0=0$

Or, maybe more understandable.

Suppose that $f(x)\ne 0$ for some $x\in(a,b)$ . and let $\varepsilon=\frac{|f(x)|}{2}>0$ . Since $f$ is continuous we should be able to find some $B_{\delta}(x)$ such that $f\left(B_{\delta}(x)\right)\subseteq B_{\varepsilon}(f(x))\subseteq \mathbb{R}^+$ . But, no matter what $B_{\delta}(x)$ you pick there will be some $q\in\mathbb{Q}$ such that $q\in B_{\delta}(x)$ and so $f(q)=0\in f\left(B_{\delta}(x)\right)\notin (0,\infty)$

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