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The GRE Practice book includes this problem (#42).

Let $\mathbb{Z}^+$ be the set of positive integers, and $d$ be a metric defined on $\mathbb{Z}^+$ by: $d(m, n) = \begin{cases} 0 & m=n \\ 1 & m \neq n. \end{cases}$

Which of the following statements are true about this metric space?

I. If $n \in \mathbb{Z}^+$, then ${n}$ is an open subset of $\mathbb{Z}^+$.

II. Every subset of $\mathbb{Z}^+$ is closed.

III. Every real-valued function on $\mathbb{Z}^+$ is continuous.

(A) None

(B) I only

(C) III only

(D) I and II only

(E) I, II, and II

My thoughts so far

Choice I) Is it talking about the set of all $n$ (eg. $\{1, 2, 3, \ldots\}$) or is it talking about any individual singleton (eg. $\{2\}$ or $\{9\}$)? I assumed the latter and decided this was false because no elements in the set $\mathbb{Z}^+$ are interior points.

Choice II) True for the same reasoning as above.

Choice III) True

The “correct” answer

According to the practice test, the answer is E, all three. I wasn't surprised that I got the answer wrong because I'm struggling with open vs closed sets when it isn't the easy cases in $\mathbb{R}^n$, but I really don't see how both choices I and II can be True. They sound obviously mutually exclusive to me.

Please help me figure out what I'm missing.

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If $n\in\mathbb{Z}^+$, then $\{n\}=B_1(n)$. Therefore, $\{n\}$ is an open set.

If $M\subset\mathbb{Z}^+$, then $M=\bigcup_{n\in M}\{n\}$. Since this expresses $M$ as an union of open sets, $M$ is open. But then $M$ is closed too, because, $\mathbb{Z}^+\setminus M$ is open.

And never forget: sets are not doors. A set may well be simultaneously open and closed and it can also be neither open nor closed.

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  • $\begingroup$ doors can be ajar! $\endgroup$ – qbert Oct 22 '17 at 21:17

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