Let $f:\mathbb{R}^2\to \mathbb{R}$ be defined as $f((s,t))=2s$. Let $A\subseteq \mathbb{R}^2$ be an open set. Does this function map $A$ to an open set? Give a counterexample if false or proof if true.
I think this is true. My proof:
Let $\varepsilon > 0$. Since $A$ is open, for every $\vec{a}\in A$, $\exists \delta > 0$ such that $B(\vec{a}, \delta)\subseteq A$, where $B(\vec{a}, \delta)=\{\vec{x}\in \mathbb{R}^2: \|\vec{x}-\vec{a}\|<\delta\}$. Since $f$ is Lipshitz with constant $c=2$, that is $|f((s,t))-f((s',t'))|= 2|s-s'|\le 2\sqrt{(s-s')^2+(t-t')^2}=2\|(s,t)-(s',t')\|=2\|\vec{x}-\vec{a}\|$, $f$ is continuous on $A$, so that $2|s-s'|<\varepsilon\iff |s-s'|<\frac\varepsilon2$. Thus, $f(B(\vec{a},\delta))=\{s\in \mathbb{R}:|s-s'|<\varepsilon/2\}=B(s',\epsilon /2)$, which is an open ball in the image of $A$ under $f$. Hence, $f(A)$ is open.
Do you think this approach might be at least approximately correct?
Now, I also think there's probably a simpler approach: Simple $f$ is Lipschitz, it is continuous, so that, for every $\varepsilon >0$, $\exists \delta > 0$ such that $f(B(\vec{a}, \delta))\subseteq B(f(\vec{a}),\varepsilon)=B(2s', \varepsilon)$. But here the ball is centred at $2s'$ instead of $s'$, and it is of a different radius.
Would appreciate your feedback.
Hint : $f(x,y) =x$ maps open square to open interval. DONE.
P.S any open set in $R^2$ can be written as Union (Possibly infinite) of open squares!