Given
$A = \{x| 1<\|x\|<3\} \cup \{(0,0)\}$
Find $\bar A$
My hunch is $\overline A = \{x| 1 \leq x \leq 3\} \cup \{0,0\}$, but my friend says I am wrong, the closure of $A$ must be $\overline A = \{x| 1 \leq x \leq 3\}$
I am curious as to why I am wrong:
my textbook (Charles Pugh real mathematic analysis) says that $\overline A$ is the set of all limit of $A$. A limit is defined as a point for which there is a sequence in $A$ converging to it. $\{0,0\}$ is a limit of $A$ because the constant sequence $(z_n) = (0,0)$ converges to it.
my friend's textbook (Munkres Topology) says that $\overline A$ is union of $A$ and the set of all limit points of $A$. $\{0,0\}$ is not a limit point, because there is a neighborhood that does not contain points in set $A$ other than $\{0,0\}$
What is the closure and who is right?
While it is true that $(0,0)$ is not a limit point of $A$, it is contained in $A$, hence must be in the closure. This doesn't contradict the definition in Munkres because he defines the closure to be the union of $A$ and the set of limit points of $A$.
By the way, I'm assuming that $A$ is being viewed as a subset of $\mathbb{R}^2$ with the standard Euclidean topology.