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Quiz 4
Let $(M,d)=(\mathbb{R}^2,d)$ where $d(x,y) = \left\{ \begin{array}{ll} 1,& \quad x \neq y \\ 0, & \quad x=y \end{array} \right.$. Let $S = \{(x,y) \colon x^2+y^2 < 1 \}$. Is $S$ open? Is $S$ closed? Explain why or why not.
Solution:
$S$ is closed
Let $(a,b)$ be a limit of $S$. This means there is a sequence $(x_n,y_n)$ in $S$ with $(x_n,y_n) \rightarrow (a,b)$. In other words, for any $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that if $n \geq N$, then $d((x_n,y_n),(a,b))<\epsilon$. Choosing $0<\epsilon<1$ in the definition of convergence means that, eventually, the sequence $(x_n,y_n)$ must be eventually constant, since the only values that $d((x_n,y_n),(a,b))$ may take are $0$ and $1$. This means there exists $N \in \mathbb{N}$ such that if $n \geq N$, then $(x_n,y_n)=(a,b) \in S$. Thus the limit of the sequence is in $S$, proving that $S$ is closed.
$S$ is open
Whether $S$ is open or not boils down to the following question: given any point $(a,b)$ in the set, can I find a small enough radius that all points in the bigger space $M$ that lie a distance less than $r$ from $(a,b)$ also lie in $S$?
Let $(a,b) \in S$ and choose $0 \lt r \lt 1$. Then the set of points $(q_1,q_2) \in S$ with $d((a,b),(q_1,q_2)) \lt r$ contains only the single point $(a,b)$ (all other points are a distance $1$ from it!). Therefore we conclude that $S$ is open!
