Answer to Question #153611 in Calculus for Arora

Question

State true or false giving proper justification for following statement:there exists a continuous function f:RtoR2 such that f(down)=(n,1/n) for all n belongs to N

Answer 1

$f:\R\rightarrow \R^2\\ f(cos(n))=(n,\frac{1}{n})$

Let f is continuous on $[-1,1]$

$\Rightarrow$ $f$ is bounded on $[-1,1]$.

i.e.

$\exists M\in \N \ni \\ \;\;\;\;\;\;\; \;|f(x)-f(1)|<M\forall x\in [-1,1]$

Now take $x=cos (M+1)$

$\therefore f(x)=(M+1,\frac{1}{M+1})\\ \Rightarrow |f(x)-f(1)|=|(M+1,\frac{1}{M+1})-(1,1)|\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\sqrt{M^2+(1-(\frac{1}{M+1})^2}\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;>\sqrt{M^2}=M\\\;\\ \therefore |f(x)-f(1)|>M$

That contradicts the bounded property.

Therefore $\pmb{\exists no\; f\ni f}$ is continuous.