Proof that [0,1] is compact (not using Heine Borel). Proof check

[fleazo]

I would like to prove [0,1], as a subset of R with the standard Euclidean topology, is compact. I do not want to use Heine Borel. I was wondering if someone could check what I've done so far. I'm having trouble wording the last part of the proof.

Claim: Let $\mathbb{R}$ have the usual Euclidean topology T. Then [0,1] is a compact subset in $\mathbb{R}$.

Proof:

Denote the set S_m as all $m \in \mathbb{R}$ such that $m < 1$ and the set [0,m] can be covered by a finite number of open sets in T.

We first show that $S_{m} \neq \emptyset$, and has an upper bound.

To show that $S_{m}$ is non-empty, note that 0 $\in S_{m}$. This can be shown by noting that if m = 0, then the corresponding set is [0,0], which is just the singleton element $\{0\}$. To find a finite covering of $\{0\}$ in T, select any $\epsilon \in \mathbb{R}$. Then $(-\epsilon,\epsilon)$ in a basis element in T, and it contains the singleton element 0. This shows that the set [0,0] can be covered by a finite number of open sets from T, $\Rightarrow 0 \in S_{m}$. Also, by assumption, we know that $\leq 1 \Rightarrow$ has an upper bound. By the completeness property of $\mathbb{R} S_{m}$ must have a least upper bound. Let's denote this as $m_{u}$.


As a next step, I'll assume that $m_{u} < 1$, and arrive at a contradiction. This will then prove that $m_{u} = 1 \Rightarrow 1 \in S_{m} \Rightarrow [0,1]$, by definition of $S_{m}$ can be covered by a finite number of open sets in T. (Since the definition of compactness is that any open cover of a set has a finite subcover, this will be sufficient to show that [0,1] is compact)


So, assume that $m_{u} < 1$. Since $m_{u} \in S_{m} \Rightarrow [0,m_{u}]$ can be covered by a finite number of open sets in T. Let $O_{a}$ be the union of all such open sets. By topology axioms, we know that $O_{a}$ is an open set itself. **This is where I am having trouble wording things.** My goal now is to essentially say that, since $m_{u} \in O_{a}$ and T is the Euclidean topology, I can find some $\delta$ where $\delta < |1-m_{u}|$ s.t. $m_{u} + \delta \in O_{a}$. This would then mean that $[0,m_{u}+\delta]$ is covered by $O_{a}$ as well, and since $m_{u}+\delta < 1 \Rightarrow m_{u}+\delta \in S_{m} \Rightarrow m_{u}$ can't be the least upper bound, which would be my contradiction.

 

[Terandol]

It seems to me there are a couple problems with this proof even before getting to the part you are having trouble phrasing. Since this is a standard proof you can find in any intro analysis textbook, it seems you want to come up with a proof on your own so I won't tell you exactly how to fix things, just which things need to be fixed.

First, it is certainly not enough to prove that a set can be covered by finitely many open sets to prove it is compact. For example, (0,1) can of course be covered by the open sets (-1,1/2) and (1/3, 2) however it is not compact. You have to start with an arbitrary open cover and prove this cover has a finite subcover. To fix this in your proof just start with an arbitrary open cover $\mathcal{C}$ from the outset and slightly redefine $S_m$ using this open cover.

Further, you assume a couple times in your proof that the least upper bound $m_u$ of the set $S_m$ is in $S_m$. This needs to be proven since in general a least upper bound of a set need not be in the set. In fact, this is a rather crucial way in which your proof does not work since by your definitions, $1\not\in S_m$ so if you prove $1= m_u$ then you also prove that $m_u\not\in S_m$. To fix this you need to slightly redefine $S_m$ again and then give a proof that $m_u\in S_m$.

Finally if you want a precise way to find the $\delta$ you are looking for at the end of your proof, just write down what a basis element containing the point $m_u <1$ would look like. Since an open set contains a basis element around every point in the set, it should be clear from this how to find a $\delta$ that works from the basis element.

 

[pasmith]

I would like to prove [0,1], as a subset of R with the standard Euclidean topology, is compact. I do not want to use Heine Borel. I was wondering if someone could check what I've done so far. I'm having trouble wording the last part of the proof.

Claim: Let $\mathbb{R}$ have the usual Euclidean topology T. Then [0,1] is a compact subset in $\mathbb{R}$.

Proof:

Denote the set S_m as all $m \in \mathbb{R}$ such that $m < 1$ and the set [0,m] can be covered by a finite number of open sets in T.

This is not the set you need. Indeed, for every $m \in (-\infty, 1)$ the set $[0,m]$ can be covered by just a single open set (the empty set if $m < 0$ and $(-1,2)$ if $m \in [0,1)$).

As an aside, if you call a set $S_m$ then you should not be using $m$ to denote an arbitrary element of that set, since the expectation is that $m$ will appear as a variable in some condition which arbitrary members of $S_m$ must satisfy, for example $\{ x \in \mathbb{R} : x < m^2 + 5m + 3\}$ or $\{ mz : z \in \mathbb{Z}\}$.

A set is compact if and only if every open cover admits a finite subcover. Therefore let $\mathcal{U}$ be an arbitrary open cover of $[0,1]$, and let $$S = \{ x \in [0,1] : [0,x]\mbox{ is covered by a finite subcollection of \mathcal{U}}\}.$$ Note that if $x \in S$ then $[0,x] \subset S$. Your aim is to show that $\sup S = 1$.