Prove that set $A=\{{(x,x)\in \mathbb R ^2 | x \in \mathbb R\}}$ is open/closed/none So I think the set $A=\{{(x,x)\in \mathbb R ^2 | x \in \mathbb R\}}$ is closed and my idea of proving it was (in general):
Consider $B=\mathbb R ^2$ \ $A$ then for every $p=(x_0,y_0) \in B$ we could take the closed point $p_0 \in A$ (that would be $(\frac {x_0+y_0} {2}, \frac {x_0+y_0} {2})$) and then
$d(p,p_0)=\frac{y_0-x_0} {\sqrt2}$.
Now we can choose $r=\frac{d}{2}$ and get $B_r(p)\subseteq B$.
Is this proof right? Am I missing something here?
Thanks.
 A: As mjw mentioned above, $A^c = P \cup P'$ of two open half planes, so $A^c$ is open, hence $A$ closed. Another way to think about this is let $x \notin A$, then the distance from this point to the line is positive, $d$. Therefore take the neighbourhood around x with radius less than $d$. Certainly this neighbourhood lies in $A^c$, therefore about every point $x\in  A^c$ you have an open neighbourhood of $x$ contained in $A^c$, so $A^c$ is open, hence $A$ closed. On the other hand $A$ is not open, seen by taking any a point on the line $A$, say the origin, then any neighbourhood $N_r$ of radius contains the intersects the x-axis at say $(\frac{r}{2},0)$ which is not on the line!
A: The set is closed because its complement (the union of two open sets (half planes)) is open.
The set is not open because if you take a point $p \in A$, any ball around $p$ is not contained in $A$.
A: Yes, it is right. Except for a small change that $d(p, p_0) =\frac{ |y_0 -x_0|}{2}$. (To ensure it is positive.)
Another way, to note this, if you know some topology, is to note that $\mathbb{R}$ is hausdorff, and thus, its diagonal which is $A$, is closed.
