Show that Set S=$\left\{ \left(x,y\right);x^{2}+y^{2}\leq1,x<1\right\} $is open or closed?
I am confused between two books
The topology problem solver says set S is neither open nor closed
Another local book Says SEt S is closed $S^{c}$=$\left\{ \left(x,y\right);x^{2}+y^{2}>1\right\} $Now prove that $S^{c}$=$\left\{ \left(x,y\right);x^{2}+y^{2}>1\right\} $ is open Let A=$\left(x_{o},y_{o}\right)$,A$\in$ $S^{c}$=$\left\{ \left(x,y\right);x^{2}+y^{2}>1\right\} $
$\delta<\sqrt{x_{o}^{2}+y_{o}^{2}}$-1 then N$_{\delta}$$\left(A\right)$$\subset S^{c}=\left\{ \left(x,y\right);x^{2}+y^{2}>1\right\} $$\Longrightarrow S^{c}=\left\{ \left(x,y\right);x^{2}+y^{2}>1\right\} is$ open $\Longrightarrow$$\left\{ \left(x,y\right);x^{2}+y^{2}\leq1\right\} $is closed
My ApproachLet $f:$$\left(x,y\right)\mapsto$$x^{2}+y^{2}$Function is continuous then$f^{-1}$$\left(1,\infty\right)$$\Longrightarrow S^{c}=\left\{ \left(x,y\right);x^{2}+y^{2}>1\right\} is$ open $\Longrightarrow$$\left\{ \left(x,y\right);x^{2}+y^{2}\leq1\right\} $is closed
I don't know which answer is correct
Edits Sorry Friends it was my mistake .I did not read the question properly Actually S is intersection of a closed and a open set That why it is neither open nor close
No, that's not the reason. The intersection of an open and closed set can be either or neither. An example is given by $[-2,2]\cap (-1,1)=(-1,1)$ and $(-2,2) \cap[-1,1]=[-1,1]$.
Now let $$S=\{(x,y)\ |\ x^2+y^2\leq 1\text{ and }x<1\}$$
$S$ is not closed: Take the sequence $(x_n,y_n)=(1-\frac{1}{n},0)$. Is is fully contained in $S$ but the limit $\lim(x_n,y_n)=(1,0)$ which does not belong to $S$.
$S$ is not open: Take $(-1,0)\in S$ and note that every open neighbourhood around it contains some point of the form $(a,0)$ with $a<-1$ which does not belong to $S$. Hence no open neighbourhood of $(-1,0)$ is fully contained in $S$ and thus $S$ is not open.