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?

  • $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ 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. $\endgroup$ – Abhay Oct 23 '19 at 9:37
  • $\begingroup$ I am also using Bolzano Wieersatrass Theorem. The problem with your argument is you may get different subsequence for different coordinates. These may not result in a subsequence of the vector sequence. Try to see what happens in two dimensions. @Abhay $\endgroup$ – Kavi Rama Murthy Oct 23 '19 at 9:41

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