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Is G open in R^2

G-X

New member
Jul 11, 2020
21
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.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,740
Looks correct.
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
433
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
Mar 10, 2012
833
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.