Proving $g:P\to\Bbb R, g(x)=3x+2$ is discontinuous

Question

Define $g:P\to\Bbb R,g(x)=3x+2$ where $P$ is the Cantor set. Show that $g$ can't be continuous under Euclidean metric in $\Bbb R$.

No information is given about the metric on $P$. I assumed it is the Euclidean metric. Then $g^{-1}((a,b))=\left(\frac{a-2}3,\frac{b-2}3\right)\cap P$ which is open in $P$. Then this map is continuous because the inverse image of every open set is open. In fact it seems $g$ will be continuous on any subspace of $\Bbb R$ since it is continuous on $\Bbb R$. Am I doing something wrong?

Answer 1

You are right and the question is wrong.

If we want to spell it out:

Let $a \in P$. Let $\epsilon > 0$. Let $\delta = \frac {\epsilon}3$.

For all $x \in P$ so that $|x-a| < \delta$ then

$|(3x + 2) - (3a+2)|= 3|x-a| < 3\delta = \epsilon$ and so $f$ is continuous at all $a \in P$.

Can't get much more plain than that.

Answer 2

Theorem If $A\subset X$ and $f\colon X\to Y$ is continuous function, then $f|A\colon A\to Y$ is continuous function.

Now, $g\colon \Bbb R\to \Bbb R$ and $g(x)=3x+2$ is definitely continuous function. Put, $$f=g|P \ \ \ \text{for all } \ \ \ x\in P$$ So, $f\colon P\to \Bbb R$ where $f(x)=3x+2$ is continuous function by Theorem.