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.


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.