A set $S \subset \mathbb{R}$ is closed if the set $S$ contains all of its limit points.
A point $\boldsymbol{x}$ is a limit point of a set $S$ if there is a sequence $ \{\boldsymbol{x_{n}}\}_{n = 1}^{\infty}$ such that $\boldsymbol{x_{n}} \in S$
Show that the hypercube $[a,b]^n$ is closed.
Attempt:
So I successfully showed that $[a,b]$ is closed, but I can't just extend that idea because there $\mathbb{R}^n$ is not ordered. With that being said:
Proof:
Let $\boldsymbol{x}$ be a limit point of $[a,b]^n$.
This means there exists a sequence $\boldsymbol{x_{n}}$ that converges to $\boldsymbol{x}$. This implies that the sequence $\boldsymbol{x_{n}}$ is bounded. So by Bolzano Weirstrauss there is a convergent subsequence. As such there is a convergent subsequence for each component. Taking convergent sub-subsequnces of each component by induction I will show that every limit point of hypercube $[a,b]^n$ is in hypercube $[a,b]^n$.
Note: I feel that I am not being explicit enough in my explanation or perhaps missing something to make the proof sound. What suggestions could be given to improve upon this?
Suggestion: Suppose $p_j=(x_{1,j},...,x_{n,j})\in [a,b]^n$ for each $j\in \Bbb N,$ such that $(p_j)_{j\in \Bbb N}$ is convergent in $\Bbb R^n$ to $q=(q_1,...,q_n).$
Since $$|p_{j,m}-q_m|\leq \|p_j-q\|$$ for $1\leq m \leq n$ and since $\|p_j-q\| \to 0$ as $j\to \infty,$ therefore each sequence $(p_{j,m})_{j\in \Bbb N}$ must converge to $q_m.$
Now $p_{j,m}\in [a,b]$ so $q_m\in [a,b]$ for each $m.$