Can someone please work this out for me/give hints.
Give an example of a continuous function $f: \mathbb R^2 \rightarrow \mathbb R$, along with a closed set $K \subset \mathbb R^2$ such that $f(K)$ is not closed.
This implies that $K$ is not bounded as it would be compact otherwise. Check this explicitly by giving a sequence of points $\{x_n\}$ in $K$ with no Cauchy subsequence.
Thanks!
Let $K=\{(x,y) \in \mathbb R^{2}: x>0,y>0,xy=1\}$. Let $f(x,y)=x$. Then $f(K)=(0,\infty)$ which is not closed. To show that K is closed suppose $\{(x_n,y_n)\}$ is a sequence in K converging to some point $(x,y) \in \mathbb R^{2}$. Then $x_n y_n =1$ for all n and $x_n \to x$,$y_n \to y$. taking limits on both sides we get $xy=1$. Note that $x_n >0$ for all n and $x_n \to x$ give $x \geq 0$. However $x=0$ is ruled out by the condition $xy=1$. Similarly $y>0$. Thus K is closed. $\{(n,1/n)\}$ is a sequence in K with no convergent subsequence.