Proving compactness and connectedness of a set

Question

$A=\left \{ \left ( x,\frac{1}{x}\right );1\leq x\leq 2 \right \}$

I proved that $A$ is a closed and bounded subset of $\mathbb{R}^2$, then it's compact (correct ?)

And is $A$ connected because it's homeomorphic to a closed interval ? If yes how would I prove it ?

Answer 1

Yes, you are correct. More precisely, define

$$f: [1,2] \to \mathbb{R}^2: x \mapsto \left(x, \frac{1}{x}\right)$$

Then clearly $f$ is continuous. Since $[1,2]$ is compact and connected, so is the continuous image $A= f([1,2])$.

Answer 2

The map$$\begin{array}{rccc}f\colon&[1,2]&\longrightarrow&A\\&x&\mapsto&\left(x,\frac1x\right)\end{array}$$is a homeomorphism between $[1,2]$ and $A$; its inverse is defined by $f^{-1}(x,y)=x$. This proves that $A$ is both compact and connected.