Every closed and bounded subset of $(R^n,||\cdot||_p)$ is compact $1\leq p \leq \infty$

Every closed and bounded subset of $$(R^n,||\cdot||_p)$$ is compact $$1\leq p \leq \infty$$

I will show that every closed and bounded subset of the above metric space is sequentially compact, and hence compact.

Let $$(x_k)_k$$ be sequence in $$F\subseteq (R^n,|| \cdot||_p), \; 1\leq p \leq \infty$$. $$F$$ is closed and bounded subset of $$R^n$$

then, $$(x_k^i)_k \; 1\leq i \leq n$$ is a bounded sequence of $$R$$. So by Bolzano Weierstrass Theorem, there exists subsequences such that $$(x_{k_t}^i)_t \to y_i\,; \; 1\leq i \leq n$$

then, $$(x_{k_t})\to x_0$$

where $$x_0 = (y_1,y_2\ldots,y_n)\in R^n$$ because convergence in $$(R^n,||\cdot||)$$ happens only if coordinate wise convergence happens.

But since $$F$$ is closed, and $$(x_{k_t})$$ lies in $$F$$, so $$x_0\in F$$

hence, we get a subsequence of $$(x_k)_k$$ converging to a point in $$F$$

So $$F$$ is sequentially compact.

Is this correct?

• Hint: all norms on a finite-dimensional vector space are equivalent, in other words they generate the same topology. And compactness is a topological property. – Math1000 Oct 23 '19 at 16:22

In your argument you are picking some subsequence for each coordinate. This need not give you a subsequence of the sequence of vectors $$(x_k)$$. Instead, you can argue as follows. Pick a subsequence along which the first coordinates converge. The look at the second coordinates along this subsequence and extract a a further sub sequences along with the second coordinated converge. And so on.
• I don't get this. Can you please elaborate. How do I pick a subsequence along which the first coordinates converge. Here $(x_k)_k$ need not be convergent in first place. – Abhay Oct 23 '19 at 9:37