$K \subset \mathbb{R}^n$ is compact iff it is closed and bounded I want to prove $\Leftarrow$.
I just need one more step, now. If I find a compact set that contains K, then it will finish the proof. Rudin take a k-cell to do that, but before it he proves that every k-cell is compact, and I think I can take a more simple compact to do this step. I has thinked in the closed ball, but don't know how to prove that it is compact. Can anyone help me to find this set, if it really exists? Thanks.
 A: One often needs a starting point, on which we do "laborious" work, and from which facts stem smoothly. The starting point often used for the facts about topology of $\mathbb{R}^n$ is the $k-$cell. There are several reasons for this: 


*

*It is simple.

*It has a natural "combinatorial" presentation. Look up, for instance, the proof that the $k$-cell is compact and you will see that the way the subdivision works is neat.

*It "dominates" the topology of $\mathbb{R}^n$, in the sense that a lot of sets can be seen from doing something topologically to the $k$-cell.


etc.
Rudin aggrees with this, and proves that the $k$-cell is compact first. The $k$-cell is not only a nice starting point for compactness, but also for connectedness. From the fact that the $k$-cell is connected we can infer a lot of facts about the topology of $\mathbb{R}^n$ (in fact, about many other things too). 
But there is no free meal. We need to do work to arrive at those facts. And we choose to do it for the $k$-cell because, among other reasons, it is simple and useful.
Now, to answer the question directly: The $k$-cell is simple enough.
