Is G open in R^2

G-X

New member
The set G = {$$\displaystyle (x, y) \in R^{2} : (x, y) \neq (1, 0)$$} is an open set in $$\displaystyle R^{2}$$

Proof 2: We know a set X is defined to be closed if and only if its complement is open. Let $$\displaystyle J$$ = $$\displaystyle R^{2} / G$$ which implies $$\displaystyle J$$ is a single point (1, 0) in $$\displaystyle R^{2}$$. We can prove $$\displaystyle G$$ is open if we prove J is closed.

We can use the definition $$\displaystyle \bar A = A \cup A'$$ where $$\displaystyle \bar A$$ is the closure of $$\displaystyle A$$ and $$\displaystyle A'$$ is the set of all limit points of $$\displaystyle A$$. This means that the closure of a set $$\displaystyle A$$ is the union of $$\displaystyle A$$ with its boundary points. The closure of a set $$\displaystyle A$$ is, also, the smallest closed set containing $$\displaystyle A$$.

This implies that $$\displaystyle \bar A$$ is closed and we can show J is closed by proving that it is equal to its closure. Let us set $$\displaystyle \bar J = J \cup J'$$. By definition of a limit point, let $$\displaystyle b \in S$$. A point $$\displaystyle x \in S$$, $$\displaystyle x \neq b$$ is a limit point of $$\displaystyle b$$ if and only if every open neighborhood of $$\displaystyle x$$ contains $$\displaystyle b$$. Except $$\displaystyle J$$ is a singular point and has no other elements $$\displaystyle x$$. This implies that $$\displaystyle J$$ has no limit points and $$\displaystyle J' = \{\}$$, the null set.

This means that $$\displaystyle \bar J = J \cup J' = J \cup \{\} = J$$. We now know that J is equal to its closure and thus J is closed. In turn, this means that G is open.

Staff member
Looks correct.

Country Boy

Well-known member
MHB Math Helper
That is correct but I would have been inclined to a direct proof.

Let (x, y) be a point in the set G. Then x and y are not both 0 so the distance from (x, y) to (0, 0), $\sqrt{x^2+ y^2}= \delta$ is positive. So $\frac{\delta}{2}$ is also positive. The points in the neighborhood, $N\left((x,y), \frac{\delta}{2}\right)$, have distance from (0, 0) at least $\delta- \frac{\delta}{2}= \frac{\delta}{2}$ so the neighborhood is a subset of G. That is, given any point in G there exist a neighborhood of that point that is a subset of G. Therefore, G is open.

caffeinemachine

Well-known member
MHB Math Scholar
I just want to add one other approach, which I believe is worth learning and it is very easy to apply otherwise. Define a map $f:R^2\to R$ by writing $f(x, y) = (x-1)^2 + y^2$. This map takes the value $0$ only at $(x, y) = (1, 0)$. Thus $G=f^{-1}(R\setminus \{0\})$. Since $f$ is continuous, it suffices to show that $R\setminus \{0\}$ is open. This is easy to show, and certainly easier that showing $G$ is open.