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.