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Show that a subset of $\mathbb{R}$ is compact iff it is closed and bounded.

By definition of compact, a set $S\subset\mathbb{R}$ is compact if every open covering of $S$ has a finite subcovering. So for a given open covering {$U_{n}$}, there exista a finite covering $U_1,...,U_n$ of {$U_{n}$} s.t. $S\subset U_1\cup...\cup U_n$.

Now that I have all this out of the way, I'm not exactly sure where to start.

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@Emir: If you plan on posting several queries, please consider registering for the site; this will keep all your questions under your user name, and other goodies. – Arturo Magidin Jan 24 '12 at 6:25
up vote 6 down vote accepted

Proof of this theorem is not difficult, but long and not so funny. I leave you some hints.

  1. Proving compact $\Rightarrow$ closed and bounded.
    1. Prove that, in a Hausdorff space, every compact subspace is closed. This is not so trivial, I think. Let be $Y$ a compact subspace of a topological space $(X,\tau)$. We prove that $X\setminus Y$ is open. Let be $x_0 \in X\setminus Y$ a generic but fixed point; by separation axiom, there exist (open) neighborhoods $U_y$ for every $y\in Y$ in $X$ and $V_y$ of $x_0$ in $X$ with the propriety that $U_y\cap V_y=\varnothing$. Now, the family $\mathfrak{U}=\{U_y\,|\,y\in Y\}$ is a cover of $Y$; so there is a finite number $k$ of elements $y_1,\ldots,y_k$ such that $\{U_{y_i}\,|\,i=1,\ldots,k\}$ covers $Y$ and such that all elements are disjoint from the respective Hausdorff neighborhood $V_{y_i}$. Then we consider the element $V=V_{y_1}\cap\ldots \cap V_{y_k}$, that is an open neighborhood of $x_0$ in $X\setminus Y$. We can repeat the costruction for every $x_0\in X\setminus Y$, so the thesis.
    2. As obvious corollary, in a metric space every compact subsace is closed. So in $\mathbb{R}^n$.
    3. Show that every compact subspace in a metric space $(X,d)$ is necessarily bounded with respect to metric $d$ (this is easy).
  2. Proving that closed and bounded $\Rightarrow$ compact.
    1. A first step can be the proof that being bounded for a $Y\subseteq \mathbb{R}^n$ does equal to the existance of a closed pluri-rectangle $Q=[a_1 ,b_1]\times\ldots\times [a_n,b_n]$ that contains $Y$.
    2. By a simple version of Tychonoff theorem, that you can prove quite easily, the finite product of compact spaces is compact in the ambient product space. So does the pluri-rectangle as product of compact spaces (in $\mathbb{R}$).
    3. You know that $Y$ is closed in $\mathbb{R}^n$, so is also closed in the compact subspace $Q$.
    4. A closed subspace in a compact space is already compact. This fact has a very simple and straight proof.
    5. So you proved that $Y$ is compact.

A very boring part that I switched is the proof that every closed interval in $\mathbb{R}$ is compact. I think you should have seen that.

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You decided to give the proof of the more general version in R^n instead of the one on the real line.I wouldn't object,but I'm not sure the student's comfortable working in abstract metric spaces.It'll depend on what kind of first analysis course they've had.Nice job guiding him/her through the proof with the hints. – Mathemagician1234 Jan 24 '12 at 20:01

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