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I have the following problem:

Let the constraint set be $D=\{(x,y) \in \Re^2 \;| x^2 + y^2=1\}$, show that the $f(x,y)=x^2-y^2$ is a continuous function on $D$ and also that $D$ is compact.

Showing that $\lim_{(x,y)\rightarrow(a,b)}f(x,y)$ is equal to $f(a,b)$ suffices to prove the continuity?

To compactness is obvious that $D$ is bounded, by $B(0,2)$ for instance. But how to show that is closed?

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  • $\begingroup$ For your Q on proving the continuity: YES. $\endgroup$ – DanielWainfleet Apr 4 '17 at 23:14
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On D being closed: Two methods:

$(\;(x_n,y_n)\;)_n$ converges to $(x,y) \implies (\;(x-x_n)^2+(y-y_n)^2\;)_n$ converges to $0\implies (\lim_n x_n=x$ and $\lim_ny_n=y)\implies (\lim_n x_n^2=x^2$ and $\lim_ny_n^2=y^2)$.

If $(x_n,y_n)\in D$ and $(\;(x_n,y_n)\;)_n$ converges to $(x,y)$ then $$|x^2+y^2-1|=|(x_n^2+y_n^2-1)+(x^2-x_n^2)+(y^2-y_n)^2 |=$$ $$=|(x^2-x_n^2)+(y-y_n^2)|\leq |(x^2-x_n^2)|+|(y^2-y_n^2)|$$ which is $<r$ for any $r>0$ if $n$ is big enough. So $|x^2+y^2-1|$ is less than any $r>0$, so $x^2+y^2-1=0$. So $(x,y)\in D.$

Another method: Show that the complement of $D$ is open. With $\|(x,y)\|=\sqrt {x^2+y^2}$ we have the Triangle Inequality: For $u,v\in \mathbb R^2$ we have $\|u+v\|\leq \|u\|+\|v\|.$

If $u\in \mathbb R^2$ with $\|u\|=1+s\ne 1$ then the open ball $B(u,|s|/2)$ is disjoint from $D$ because

(i). If $s<0$ then $$v\in B(u,|s|/2)\implies \|v\|=\|(v-u)+u\|\leq \|v-u\|+\|u\|<$$ $$<|s|/2+(1+s)=|s|/2+(1-|s|)=1-|s|/2< 1.$$

(ii). If $s>0$ then $$v\in B(u,|s|/2)\implies \|v\|=\|u-(u-v)\|\geq \|u\|-\|u-v\|>$$ $$>(1+s)-|s|/2=(1+|s|)-|s|/2=1+|s|/2>1.$$

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