# Prove/disprove that the set is open or closed

Have a set, $$X=\left\{\left(\frac{1}{k}cosk,\frac{1}{k}sink\right):\:k\ge 1\right\}$$. It is a subset of $$\mathbb{R}^2$$.

Prove or disprove that $$X$$ is an open set. Prove or disprove that $$X$$ is a closed set. Then, determine $$Int(X)$$ and $$Cl(X)$$.

This is a practice question. There are many practice ones given like this, and I'm trying to figure out how to approach and solve this type of problem.

I have the definitions that I try to start with:

A set $$X\subseteq\mathbb{R}^n$$ is open if $$\forall x\in X, \exists\:r>0$$ such that $$B(x,r)\subseteq\:X$$.

A set $$X\subseteq\:\mathbb{R}^n$$ is closed if $$X^C$$ is open. $$\:\:$$ ($$X^c$$ is complement of $$X$$).

A set $$X\subseteq\:\mathbb{R}^n$$ is closed iff every sequence in $$X$$ that converges (to some element $$\mathbb{R}^n$$) has its limit in $$X$$.

Here are the approaches I have been thinking about for open:

let $$x$$ be a point in the set, and then show that there's a ball centered around it that is entirely within the set. Or given $$x$$ an element of the set, I need to figure out how small the radius must be of the ball around $$x$$ so that the ball lies within the bounds.

For closed, to prove it I need to show the complement is open. But, I know that a set doesn't need to be open or closed, and a set can be both open and closed, so maybe it's best I don't use an argument of “suppose the set is not open." Otherwise, I can show that the set contains all its limit points: so for some $$x$$, the limit of $$x$$ will be contained in $$X$$.

• And so … is it open? Commented Mar 9, 2022 at 2:15
• $k$ is a positive integer, right? Commented Mar 9, 2022 at 2:27
• No, you’re not thinking right. The set is open if and only if you can wiggle every point in it and stay within the set. Taking limits probably has more to do with closedness. Commented Mar 9, 2022 at 2:41
• No, for every $x$ in the set, an entire open ball around $x$ must be contained in the set. Commented Mar 9, 2022 at 2:49
• Yes, much better! Keep learning. Commented Mar 9, 2022 at 3:08

## 1 Answer

The sequence $$x_n\rightarrow O$$ converges to the origin $$O=(0,0)$$ which does not belong to the set $$X$$. By your third definition this means that $$X$$ is not closed.

On the other hand, a sequence $$y_n=(1+\frac{1}{n},0)$$ is contained in $$X^c$$ (notice $$|x_n|\le 1$$, while $$|y_n|>1$$). Yet, $$y_n\rightarrow x_1$$, which means $$X^c$$ is not closed, so $$X$$ is not open.

So, $$X$$ is neither open nor closed.

• i don't completely understand how $X$ is not open (sorry I'm new to this stuff) Commented Mar 9, 2022 at 3:08
• Did you show that the complement is not closed, which implies that the set is not open? is this a proof by some contrapositive or converse? I'm a bit lost with how it works Commented Mar 9, 2022 at 3:18
• @eddie . Some texts define open and closed in ways that do not make it obvious that a set is open iff its complement is closed. Commented Mar 9, 2022 at 3:52
• @eddie, yes your summary is exactly right. $X$ is open implies $X^c$ is closed (by def.); then since $X^c$ is not closed we get that $X$ is not open (by contrapositive). Commented Mar 9, 2022 at 8:37
• @eddie you can show that there is no ball around $(cos1, sin1)$ which is entirely contained in $X$ (it's very similar to what I did in the answer). So I think I agree with your question. But I do not understand your notation, it looks like you are adding vectors ($(cos1, sin1)$) and scalars $r$. Commented Mar 9, 2022 at 19:05