Prove that if Y is a closed and bounded subset of $R^{N}$ then Y is complete and totally bounded (pre-compact).
I was able to prove that Y is complete by showing that a Cauchy sequence in Y has a converging subsequence in Y and using that to say that the sequence also converges to the same value in Y. However, I am not sure about the totally bounded part. By definition, I am supposed to show that Y has a finite cover where the diameter of the sets in the cover can be made arbitrarily small.
Best Answer
If $Y$ is closed and bounded, by the Heine-Borel theorem, it is compact.
Then since $\mathbb{R}^N$ is a finite dimensional vector space, a compact set is sequentially compact, so every sequence in $Y$ has a convergent subsequence and limits are unique, so every convergent sequence converges to an element of the set. $Y$ is complete since every convergent sequence converges to an element of the set, so every cauchy sequence converges.
Since $Y$ is bounded, take the $\delta >0$ for which $Y \subset B_{\delta}(0) = \{x\in \mathbb{R}^N: ||x|| < \delta \}$. Let $\varepsilon>0$ be given. Imagine the $k-$cell that encloses $Y$ as in the proof of the Heine-Borel theorem with diameter $\delta+1$. Subdivide the $k-$cell so that the volume of each element of the cell is less than $\varepsilon$. So in $\mathbb{R}^2$, you need $k$ large enough that $ \frac{(1+\delta)^2}{2^k}< \varepsilon$, and in $\mathbb{R}^N$, it would be something like $ \frac{(1+\delta)^N}{2^{k}} < \varepsilon$; scan the proof of the Heine-Borel theorem to get the details, I have had too much vodka to remember it exactly and I don't have a pad of paper to sketch it out.
This coronavirus thing is killing me.