# Is the set $A = \{1,1/2,1/3,\ldots,1/n,\ldots\}$ open or closed

I don't think it is closed because $$0$$ is a limit point of the set (by the Archimedean principle) but $$0$$ is not in $$A$$ but $$A$$ is clearly not open, does than mean it's not open nor closed. Is that even possible?

• Yes, it is possible and you already have a complete answer. Jan 27, 2021 at 8:20
• That last sentence reminds me of this old video. More seriously, the only necessary relationship between openness and closedness is that if a set is open then its complement is closed, and voice versa. Jan 27, 2021 at 8:21
• consider $$(0,1]$$ Jan 27, 2021 at 8:24
• @Arthur Fantastic: I feel the Führer's pain. Jan 27, 2021 at 8:29
• Sets are not doors. Jan 27, 2021 at 8:35

Yes, that is deffinitely possible. A set can be both closed, open, open and closed and neither.

The only relation between the two is that the complement of a set is closed if and only if that set is open, and the complement of a set is open if and only if that set is closed.

This can be checked as a theorem in Abbott's book "A set O is open if and only if $$O^c$$ is closed. Likewise, a set $$F$$ is closed if and only if $$F^c$$ is open."

• Are you sure about your last sentence? If $A^c$ is closed, then $A$ is open, and if $A^c$ is open, then $A$ is closed. But since $A$ is neither, does it not then follow that $A^c$ is also neither closed nor open? These things confuse me. Jan 27, 2021 at 8:30
• $\mathbb{R}$ is connected so your last statement cannot be true. Or simply a one-step contradiction: if $A^c$ is open then $A$ is closed. Jan 27, 2021 at 8:32
• I am rather sure, at least that is what my lecturer taught me. I am looking up a citation in our course book right now. Jan 27, 2021 at 8:32
• The complement of a clopen set is clopen, by your second paragraph. Jan 27, 2021 at 8:34
• @Yourong'DZR'Zang ah yes, I see, I was wrong. Jan 27, 2021 at 9:27

(Unlike most posts: the indented details in this post are extra details you don't have to read. They just go into more detail. The main points are in the parts that are not indented.)

Your argument that it is not closed is exactly correct.

[$$0$$ is limit point as for any $$\epsilon > 0$$ and $$n > \frac 1\epsilon$$ we have $$\frac 1n \in N_{\epsilon}(0)$$. And $$0\not \in A$$.]

Your statement that it is "clearly not open" is true but not good enough. You will have to give a more thorough proof/argument. Buy yes it is not open.

[Easy enough to prove $$1\in A$$ and for any $$\epsilon > 0$$ then let $$w$$ be such that $$\max(1-\epsilon, \frac 12) < w < 1$$ then $$w \in N_{\epsilon}(1)$$ but $$w\not \in A$$ so $$1$$ is not an interior point so $$A$$ is not open.]

So, yes, it is neither open nor closed. Not only is that possible it is very common. The definition of open and the definition of closed are not mutually exhaustive so it is theoretically possible to have both i) not all its limit points contained in it and ii) not all of its points being limit points.

A much simpler counter example is $$[1,2)$$. It's not closed because $$2$$ is a limit point in it. And it's not open because $$1$$ is not an interior point. If you think about it, a limit point not being in it, in no way will force every point (including those that have nothing to do with the missing limit point) to be an interior point.

"open" and "closed" despite the english meaning are not an "either/or" situation.

Nor are they mutually exclusive. A set can be both open and closed. $$\emptyset$$ is both, and in $$\mathbb R$$ with the Euclidean/"usually" metric $$\mathbb R$$ is too. Those are the only examples with $$\mathbb R$$ and the Euclidean metric but there are other metrics where open and closed sets are more common.

(For example the discrete metric where $$d(x,y) =\begin{cases} 0& x=y\\1& x\ne y\end{cases}$$ is a metric where EVERY set is both open and closed. [A set $$A$$ is open because if $$x\in A$$ then $$N_{1}(x)=\{w| d(w,x)< 1\} = \{w|d(w,x)=0\} = \{w|w=x\} = \{x\}\subset A$$. So all points of $$A$$ are interior points. A set $$A$$ is closed because if $$w$$ is a limit point of $$A$$ then $$N_1(w) = \{x|d(x,y)< 1\}=\{x|d(w,x)=0\} =\{x|w=x\} =\{w\}$$ must contain a element of $$A$$. But $$N_1(w)$$ contains only one point at all. So that point $$w\in A$$. So the limit point is in $$A$$.])

so "open" does not mean "not closed" and vice versa, and both open and closed and neither open nor closed are possible.

....

But, however, (as if this post isn't long enough), it is a theorem that one connection between a set being open and a set being close is that: $$A$$ is open $$\iff$$ $$A^c$$ is closed.

But as for $$A$$ being open if or only if $$A$$ is or is not closed... there's nothing that can be said about that. All cases are possible: A set $$A$$ can be opened; closed; neither; or both.

$$A=\{1,\frac{1}{2},\frac{1}{3},...\}$$

I will now prove to you that $$A$$ is neither open nor closed.

In order for a set $$X$$ to be open, for all $$x \in X$$, $$x \in X^o$$. In other words, we need each $$x \in X$$ to have an $$\varepsilon$$-neighborhood such that $$(x-\varepsilon,x+\varepsilon) \in X.$$

Counterexample: Consider $$a \in A, a=1.$$ Surely, you would agree that $$(1-\varepsilon,1+\varepsilon) \not\in A.$$ Thus, we have found $$a \in A$$ such that $$(a-\varepsilon,a+\varepsilon) \not\in A$$. Thus, $$A$$ is not open.

In order for a set to be closed, it must contain all of its accumulation points. In other words, $$\bar A = A.$$

Counterexample: Consider $$a \in A, a = 0.$$ Notice that $$0 \in \bar A$$ but $$0 \not\in A.$$ Thus, $$\bar A = [0,1]$$ and $$A = (0,1].$$ Since $$[0,1] \neq (0,1],$$ it is true that $$\bar A \neq A.$$ Thus, A is not closed.

Now, we have shown that $$A = \{1,\frac{1}{2},\frac{1}{3},...\}$$ is neither open nor closed.

• " is essentially the same as saying A=(0,1]" What????? Jan 27, 2021 at 18:23