Compactness and the Limit

Question

Prove that the set

$Y=\{0\} \cup\left\{\frac{\cos k}{k^{3}} \mid k \in \mathbb{N}\right\}$ is a compact subset of $R$.

From the definition of compactness, it is sufficient to show the closedness and boundedness of the given set $Y$. Using the fact that the $cos$ function is bounded above for any $k \in \mathbb{N}$, we can easily deduce the boundedness of $Y$ (and if not mistaken, using the squeeze theorem implies the limit of $\frac{cos k}{k^{3}}$ is zero). However, in order to show the closedness, it will be enough to prove all boundary points of $Y$ are in the set itself - intuitively, there seems to be only one boundary point - $\{0\}$, but I can not prove mathematically why this is the case.

Answer 1

Let us prove a general fact: Let $(x_n)$ be sequence in a $\mathbb R$ which converges to a point $a=x$. Then $A=\{x\} \cup \{x_1,x_2,...\}$ is compact.

Proof: Let $(U_n)$ be an open cover of $A$. Then $x \in U_k$ for some $k$. Since $x_n \to x$ we have $x_n \in U_k$ for all $n$ sufficiently large, say for all $n \geq m$. Pick $k_1,k_2,...,k_{m-1}$ such that $x_i \in U_{k_i}$ for $ 1 \leq i <m$. Then $U_k$ together with $U_{k_i}: 1 \leq i <m$ gives a finite subcover.

Alternatively, you can show that the only limit point of $A$ is $x$, so $A$ contains all its limit points, making $A$ closed and bounded.

Answer 2

Your set is actually completely made of boundary points: every neighborhood of each point contains points which are not in $Y$. Your intuition rightfully suggested you the origin is a special point in the set but it's not the only boundary point, rather it's the only accumulation point: every neighborhood of the origin contains points of $Y$ other than $0$.