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## Main Question or Discussion Point

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

Proof:

Denote the set S

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

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

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.

Claim: Let [itex]\mathbb{R}[/itex] have the usual Euclidean topology

**T**. Then [0,1] is a compact subset in [itex]\mathbb{R}[/itex].Proof:

Denote the set S

_{m}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.