Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Sep 13, 2014 #1
    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 [itex]\mathbb{R}[/itex] have the usual Euclidean topology T. Then [0,1] is a compact subset in [itex]\mathbb{R}[/itex].


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

    We first show that [itex]S_{m} \neq \emptyset[/itex], and has an upper bound.

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

    As a next step, I'll assume that [itex]m_{u} < 1[/itex], and arrive at a contradiction. This will then prove that [itex]m_{u} = 1 \Rightarrow 1 \in S_{m} \Rightarrow [0,1][/itex], by definition of [itex]S_{m}[/itex] 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 [itex]m_{u} < 1[/itex]. Since [itex]m_{u} \in S_{m} \Rightarrow [0,m_{u}][/itex] can be covered by a finite number of open sets in T. Let [itex]O_{a}[/itex] be the union of all such open sets. By topology axioms, we know that [itex]O_{a}[/itex] is an open set itself. **This is where I am having trouble wording things.** My goal now is to essentially say that, since [itex]m_{u} \in O_{a}[/itex] and T is the Euclidean topology, I can find some [itex] \delta[/itex] where [itex] \delta < |1-m_{u}|[/itex] s.t. [itex]m_{u} + \delta \in O_{a}[/itex]. This would then mean that [itex][0,m_{u}+\delta][/itex] is covered by [itex]O_{a}[/itex] as well, and since [itex]m_{u}+\delta < 1 \Rightarrow m_{u}+\delta \in S_{m} \Rightarrow m_{u} [/itex] can't be the least upper bound, which would be my contradiction.
  2. jcsd
  3. Sep 13, 2014 #2
    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 [itex] \mathcal{C} [/itex] from the outset and slightly redefine [itex] S_m [/itex] using this open cover.

    Further, you assume a couple times in your proof that the least upper bound [itex] m_u[/itex] of the set [itex] S_m [/itex] is in [itex] S_m [/itex]. 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, [itex] 1\not\in S_m [/itex] so if you prove [itex] 1= m_u [/itex] then you also prove that [itex] m_u\not\in S_m [/itex]. To fix this you need to slightly redefine [itex] S_m [/itex] again and then give a proof that [itex] m_u\in S_m [/itex].

    Finally if you want a precise way to find the [itex] \delta [/itex] you are looking for at the end of your proof, just write down what a basis element containing the point [itex] m_u <1[/itex] 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 [itex] \delta [/itex] that works from the basis element.
  4. Oct 2, 2014 #3


    User Avatar
    Homework Helper

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

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

    A set is compact if and only if every open cover admits a finite subcover. Therefore let [itex]\mathcal{U}[/itex] be an arbitrary open cover of [itex][0,1][/itex], and let [tex]S = \{ x \in [0,1] : [0,x]\mbox{ is covered by a finite subcollection of $\mathcal{U}$}\}.[/tex] Note that if [itex]x \in S[/itex] then [itex][0,x] \subset S[/itex]. Your aim is to show that [itex]\sup S = 1[/itex].
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook