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 $X$.
Best 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.