I'm teaching myself topology using a text I found online. Right now I'm reviewing "Metrics." Please let me know if my answers are correct, and If my reasoning is accurate and complete.
I think (c)and (d) are correct, but my reason could be better stated. (not sure how)
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In each part below, a subset $S$ of a metric space $M$ is given. In each case decide whether $S$ is open, closed, both or neither.
(a)$M = \mathbb{R},$ and $S = [0,1)$
Neither. S is not open because the point {0} cannot have an open ball around it contained in $S$. S is not closed because $M-S$ has the point {1}, and it cannot have an open ball around it contained in $M-S$.
(b)$M = \mathbb{R},$ and $S = \mathbb{N}$
Closed. $M-S = …(0,1) \cup (1,2) \cup (2,3)…$ So $M-S$ is a union of arbitrarily many open subsets of M, and thus an open subset of M. And since the complement of S is open, S is closed. Not open because the point {1}, for example, can not have a ball around it contained in $S$
(c)$M = \mathbb{Z},$ and $S = \mathbb{N}$
Both. $B_{0.5}(3)=\{3\}$ is open, and $\overline B_{0.5}(3)=\{3\}$ is closed. Thus S is both, since every subset is a finite union of singletons, and finite unions preserve both openness and closedness.
(d)$M = \mathbb{R}^2,$ and $S$ is the set of point with rational coordinates
Both. Rational numbers are countably infinite, and thus a infinite union of singletons. Ergo, the argument for (c) applies.
(e)$M = \mathbb{R}^2,$ and $S$ is the unit disk $\{ (x,y) \in \mathbb{R}^2:x^2+y^2<1 \}$.
Open. This is a 2 dimensional open ball, and an open ball is an open set. Not Closed because $(1,0) \in M-S$, and $(1,0)$ cannot have a ball around it contained in $M-S$.
(f)$M = \mathbb{R}^3,$ and $S$ is the unit disk $\{ (x,y,z) \in \mathbb{R}^3: z=0 $ and $ x^2+y^2<1 \}$.
Neither. Not open, because if a ball is formed at any point on the unit disk, it will contain points in M that are not on the disk, thus it does not contain a open ball around each of its points. Not closed, because the compliment of S is \mathbb{R}^3 with the disk removed, thus it contains the unit circle ($ z=0 $ and $ x^2+y^2=1$); The points on the unit circle are boundary points, and don't have open balls.
(g)$M =\{ (x,y) \in \mathbb{R}^2 :x>0 $ and $y>0 \},$ and $S = \{ (x,y) \in M:x^2+y^2 \leq 1\}$.
Closed. $M-S$ does not contain any boundary points (they are all in S), and is therefor open, which implies that S is closed. Not open, because the points on $x^2+y^2=1 are boundary points and thus don't have open balls.
Best Answer
Jooi is correct about (d) being wrong. Your flaw when applying (c)'s logic to (d) is that in (d), $B_{0.5}(3,3) \neq \left\{(3,3)\right\}$, so points of $S$ are not open in $M$. As for closed, what are the boundary points of $S$?
Looking back at (c) I see another issue. You say every subset of $S$ is a finite union of singletons. This isn't true. But in the discrete topology (where points are open sets), every set is open, therefore every set is closed.