I try to prove that $[0,1]$ is connected.I have to follow the next steps: $R$ is complete using the construction of $R$ with Cauchy sequences $\Rightarrow$ $[0,1]$ is connected.How can I prove that?
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You can prove it using that outline, but as it's not generally true that completeness implies compactness implies connectedness you will need to use more facts here.
The first step is normally left out in early courses. Or one cheats by using some axiom about completeness. For example we used the axiom of least upper bound. This is used together with a Cauchy-sequence $x_n$ by first noting that you can construct two sequences $\inf_{j\ge n} x_j$ and $\sup_{j\ge n}x_j$, the first being increasing and the second decreasing. You then use the axiom again to see that these are convergent and that they converge to the same value which is in fact the same as $x_n$ converges to. We here uses the fact that $[0,1]$ is bounded to be allowed to use the axiom of least upper bound and also it being a closed interval to guarantee that the sequences converges within it.
For compactness you can prove this going via the property that every sequence in $[0,1]$ has a subsequence that's a Cauchy-sequence. You can do this by successively part the interval in two equal parts - at least one of them will contain infinite number of numbers from the sequence. You take this interval and selects the first number in that interval. Then you repeat with that interval, and the sequence following the selected number. Then you get a subsequence that's a Cauchy-sequence and therefore convergent.
Then you can use compactness to prove connectedness. Assume that you could decompose the interval in to disjoint open sets $A$ and $B$. Now since $A$ and $B$ are open they are themselves union of open intervals and since $[0,1]$ is covered by these intervals they could be reduced to a finite number of intervals covering $[0,1]$. You can then replace overlapping intervals with a larger interval and consequently constructing an disjoint open covering of $A$ and one for $B$ which would cover $[0,1]$. Now order the interval and consider two adjacent (open, non-overlapping) intervals. Now the lower limit of the upper interval and the upper limit of the lower are not part of the covering which contradicts the decomposition. In this part you use again that $[0,1]$ is a closed interval since the limits of the last sets to be within $[0,1]$.
skyking
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Humm... I don't think your proof of connectedness from the compacteness (last paragraph) is correct. 1.You wrote "Assume that you could decompose the interval in to disjoint open sets $A$ and $B$". Then you "construct a sequence $x_n$ in $A$ such that the distance to $B$ (being the infimum of distances to points in BB) converges to zero". How do you prove you can build such sequence, without proving or using that $[0,1]$ is connected? 2. And as you said "Now $x_n$ has a convergent subsequence which has a limit", HOWEVER such limit is in $\partial B$ (the frontier of $B$) not in $B$. – Ramiro Dec 14 '15 at 21:46
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@Ramiro $1$ has to be in either $A$ or $B$ and since $A$ and $B$ are open then so is a neighborhood $(1-\epsilon,1]$ of $1$ and since $A$ and $B$ are disjoint one of them have to not intersect that neighborhood. Therefore either $\sup A$ or $\sup B$ is no larger than $1-\epsilon$. – skyking Dec 15 '15 at 10:20
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You wrote "(such sequence exists as we can name $A$ and $B$ such that $\sup B<1$ which makes a sequence decreasing towards $\sup B$ such a sequence)". There are two problems in this approach: 1. Note that $A$ and $B$ are just disjoint open sets relative to $[0,1]$, so it may happen that $\sup A=\sup B=1$. 2. IF $\sup B<1$ then a sequence decreasing towards $\sup B$ is already convergent. No need to take subsequences, and so no need to use compactness of $[0,1]$. – Ramiro Dec 15 '15 at 10:20
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$A$ and $B$ may, each one, be the union of a countable collection of open sets relative to $[0,1]$. – Ramiro Dec 15 '15 at 10:24
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So you don't need compactness in your proof. OK. But, the other problem remains: as $A$ and $B$ may, each one, be the union of a countable collection of open sets relative to $[0,1]$, it may happen that $\sup A=\sup B=1$. You argument using $(1-\epsilon,1]$ implicitly assumes that $(1-\epsilon,1]$ is connected, but we cannot assume it when proving that $[0,1]$ is connected. – Ramiro Dec 15 '15 at 10:32
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@Ramiro It's by the topology of $\mathbb R$. A set is open if for every point $x$ in it theres an $\epsilon>0$ such that $(x-\epsilon,x+\epsilon)$ is within the set. So $(1-\epsilon,1]$ being the subset of one of the sets and the sets doesn't intersect the other set cannot intersect with $(1-\epsilon,1]$ either. – skyking Dec 15 '15 at 10:43
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Ok. So we don't need to use compactness at all. We don't need to take finite covers. $A$ and $B$ are open sets relative to $[0,1]$. So since $A$ and $B$ are disjoint and $[0,1]=A \cup B$, we can correctly claim that or $1\in A$ or $1\in B$. WLOG, let us assume $1\in A$. Then, because $A$ is an open set relative to $[0,1]$, there is $\epsilon>0$ such that $(1-\epsilon, 1] \subseteq A$. So $\sup B<1$. – Ramiro Dec 15 '15 at 11:17
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@Ramiro I've updated the answer to make it more clear that compactness is actually used. – skyking Dec 15 '15 at 11:34
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Once we have $\sup B<1$, we need to use the fact that $[0,1]$ is an INTERVAL to assure that there is a sequence in $[0,1]$ decreasing towards $\sup B$. Otherwise, such a sequence may not exist. The bottom-line is that your (original) proof did not need (nor actually use) the compactness of $[0,1]$, but it did use (in a critical way) the fact that $[0,1]$ is an interval. – Ramiro Dec 15 '15 at 11:41
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@Ramiro I thank you for your feedback, but I don't think comments on an earlier version of the answer is relevant anymore. I've updated the answer to reflect your comments. – skyking Dec 15 '15 at 11:48
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If you allow me, I would suggest that, in you new proof, you indicate where you used the fact that $[0,1]$ is an interval. It is used in a critical way. Otherwise, your new proof would also work, for instance, for $[-2,-1]\cup[0,1]$. – Ramiro Dec 15 '15 at 12:00
On the other hand $[0,1]$ is an interval and you can prove it is connected from this fact. In fact, $[0,1)$ is not complete, is not compact, but since it is an interval, it is connected as well.
– Ramiro Dec 14 '15 at 14:53