Question

Suppose we tried to apply our real analysis definitions/methods to the set of rational numbers Q. In other words, in the definitions, we only consider rational numbers. E.g., [0, 1] now means [0, 1] n Q, etc. In this setting: (a) Find an open cover of [0, 1] that contains no finite subcover. Hint: Fix an irrational number a € [0, 1] (as a subset of the reals now!) and for each (rational) qe [0, 1] look for an open interval avoiding a. (b) Prove that the function f(x) = 7212 is continuous on (0,2) but not bounded on (0, 2).

Answer 1

ANSWER :

a)

Find and open cover fo [0,1] that contains no finite subcover.

Take

a TE! € [0, 1] (as [0, 1] C|R)

Then take the cover

U1 of [0, 1] (as [0, 1] nn)

as  

U1 : Û(-1)},{ųC+2) n=1

Then for any rational

$r\epsilon \left [ 0,1 \right ]as\, \, r\neq \frac{1}{\sqrt{2}},r\, \, \epsilon\, \, \cup _{1}$

so , $\cup _{1}$ is an open conver of  $\left [ 0,1 \right ](as\left [ 0,1 \right ]\cap Q )$ .

Let is has an open subcover $\cup _{2}$ . i.e. there exist Such that

111, 112,...nik and n21, 122,...12

$\left [ 0,1 \right ]\subset \left \{ \left ( -1,\frac{1}{\sqrt{2}}-\frac{1}{n} \right ) \cup .....\cup \left ( -1,\frac{1}{\sqrt{2}}-\frac{1}{n_{1k}} \right ) \right \}$

$\cup \left \{ \left ( \frac{1}{\sqrt{2}}+\frac{1}{n_{21}},2 \right ) \cup .....\cup \left ( \frac{1}{\sqrt{2}}+\frac{1}{n_{2l}},2 \right ) \right \}$

Take  $n_{0}=max\left \{ n_{11},n_{12},...n_{1k} \right \}$

Then by density property of Q there exist$r_{1}\epsilon \left [ 0,1 \right ]\left ( as\left [ 0,1\right ] \cap Q\right )$ such that

$\frac{1}{\sqrt{2}} -\frac{1}{n_{0}}< r_{1} < \frac{1}{\sqrt{2}},Then \, \, \, r_{1}\, \, \epsilon\, \, \cup _{1}$

$Then \, \, r_{1} < \frac{1}{\sqrt{2}} \Rightarrow r_{1\notin } \left ( \frac{1}{\sqrt{2}}+\frac{1}{n_{2j}},2 \right )for\, \, 1\leq j\leq l.$

Also,

$n_{0}\geq n_{1j}\, \, for\, \, 1\leq j\leq k.$

$\Rightarrow \frac{1}{\sqrt{2}}-\frac{1}{n_{0}} \geq \frac{1}{\sqrt{2}}+\frac{1}{n_{1j}}, for\, \, 1\leq j\leq k.$

So,  $r_{1}\geq \frac{1}{\sqrt{2}}-\frac{1}{n_{0}}\Rightarrow r_{1}\geq \frac{1}{\sqrt{2}}-\frac{1}{n_{1j}}, for\, \, 1\leq j\leq k.$

$\Rightarrow r_{1}\notin \left ( -1,\frac{1}{\sqrt{2}}-\frac{1}{n_{1j}}, \right ) for\, \, 1\leq j\leq k.$

$So,r_1 \notin u_2$

So, $u_1$ contains no finite subcover .

b)

Prove that the function

$f(x)=\frac{1}{x^{2}-2}$ is continuous on [0,2] but not bounded on [0,2]
given,

$f(x)=\frac{1}{x^{2}-2},x\, \, \epsilon \left [ 0,2 \right ]\left ( as\left [ 0,2 \right ]\cap Q \right )$

Now f is discontinuous if $x^{2}-2=0$

$\Rightarrow x=\pm \sqrt 2$

But $\pm \sqrt 2 \,\,\OE[0,2](as\,\,[0,2]\bigcap Q)$

So,f is continuous on [0,2].

Take $M\epsilon |R.$   Then,by density property of Q.there exists rational number x such that

  $\sqrt2<x<min \left \{ 2,\sqrt {2+\frac{1}{M}}\right \}$   

so,

$x\epsilon [0,2]$

Then  

$\Rightarrow x^2-2<\frac{1}{M}$

$\Rightarrow x^2-2<\frac{1}{M}$

$\Rightarrow \frac{1}{x^2-2}>M\Rightarrow f(x)>M$

So,f(x) is not bounded