Prove that the set $\{(x,y): x=-y\}$ is open

Question

I tried to use the definition of an open set, but failed to succeed.

Are you sure that the question isn't about the complement of this set? — Michael Burr, Jan 21 '18 at 13:49
This is not open in the standard basis. Perhaps this topology is built on some other basis you forgot to mention. — Wintermute, Jan 21 '18 at 13:58
Possibly of interest :https://math.stackexchange.com/questions/2613163/proving-the-set-s-left-leftx-y-rightaxbyc-right-is-open/2613167#comment5398124_2613167 — Peter Szilas, Jan 21 '18 at 15:48

score 6 · Accepted Answer · edited Jan 21 '18 at 13:53

6

Of course you failed to succeed. It is not open. For instance, if $r>0$, $B\bigl((0,0),r\bigr)$ is not a subset of your set, although $(0,0)$ belongs to it.

edited Jan 21 '18 at 13:53

Jaideep Khare

19,293

answered Jan 21 '18 at 13:41

José Carlos Santos

427,504

score 1 · Answer 2 · answered Jan 21 '18 at 13:50

As @Jose Carlos Santos pointed out, the set is not open:

your set is $A=\{\; (x,y)\; :\; x+y=0\; \}=f^{-1}(\{0\})$ for $f(x,y)=x+y$. Since $\{0\}$ is closed and $f$ is continuous (for the usual topology of the plane), it follows that $f^{-1}(\{0\})$ is also closed.

In the plane $\mathbb R^2$, which is a connected set, the closed sets that are also open are $\mathbb R^2$ and the empty set $\emptyset$, so the set $A$ is not open.

Prove that the set $\{(x,y): x=-y\}$ is open

2 Answers2