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I need to prove that $[0,1]$ is connected, and one of the steps indicated is to let $c:=\sup\{t\in [0,1]:[0,t]\subseteq U\}$, where $U$ is a set which is relatively open in $[0,1]$. But is it even valid to define $c$ in such a way? This definition assumes that $\exists$ a real number which is immediately less than $\sup\{U\}$, that is immediately preceding $\sup\{U\}$, which is impossible due to the density of reals!

Would appreciate an explanation.

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    Is $U$ fixed ahead of time or is this a $c$ which depends on $U$? – 2017-01-25
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    Do you know that $0 \in U$? If $0 \in U$, then $\{t \in [0,1] | [0,t] \subset U \}$ is a perfectly well defined, non empty set with an upper bound (one), hence the $\sup$ exists. – 2017-01-25
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    The supremum of a set isn't necessarily in the set. – 2017-01-25
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    The supremum of any nonempty set $A$ of numbers in a closed set $B$ is contained in that set $B$. Ditto for infimum. – 2017-01-25
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    The last sentence above suggests a misunderstanding. – 2017-01-25
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    Your question already has a problem. You shouldn't be asking yourself if such number exists. You should be asking if the set is well defined (it is). Moreover, ask yourself if the set is empty (this depends on U). Then see which *is* the supremum (that always exists if one accepts infinite sup), or (otherwise) *if it has * a supremum – 2017-01-25
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    Yes, $0$ is assumed to be in $U$, but the set $U\cap [0,1]$ is half-open (that is, it is closed on the left, but open on the right). So $U$ must have a supremum, which is not in $U$. Then how can $c:=\sup\{t\in [0,1]:[0,t]\subseteq U\}$ exist if it is a number which comes just before the supremum, a real number? Real numbers are dense, so there cannot be a number immediately preceding another real number. – 2017-01-25
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    @EricWofsey: I think $\sup\{t\in [0,1]:[0,t]\subseteq U \}$ means the largest $t$ such that $[0,t]$ remains a subset of $U$, which is half-open on the right. So this is not the supremum of $U$, this is a number satisfying the condition that there exists a set $[0,t]$ which is the largest closed subset of $U$. But $U$ simply cannot have a largest closed subset. – 2017-01-25

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This depends on if $U$ is fixed ahead of time or not, if not, then the answer is clearly "no," as $0\not\in U$ means $c$ is the supremum of an empty set which is $-\infty$. However if you know $U$, and in particular if you know that $0\in U$ then yes it does.

Because $U$ is relatively open, it is the intersection of an open set of $\Bbb R$ and $[0,1]$ i.e. it is a disjoint union of intervals. But then if $0\in U$ we let $(a,b)\subseteq \Bbb R$ be the interval we intersected with $[0,1]$ to get the interval in this disjoint union containing $0$. Then clearly

$$c= \sup \{t: [0,t] \subseteq [0,b)\} = b.$$

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One basic fact about the real numbers is its so-called order completeness: if $A \neq \emptyset$ has an upper bound $c$ ($\forall x \in A: x \le c$) then $\sup A$ exists. It need not be a member of $A$, but it is a valid real number, that obeys the defining properties of the $\sup$, namely: $\forall x \in A: x \le \sup A$ and $\forall c: (\forall x \in A: x \le c) \rightarrow (\sup A \le c)$. This is how real numbers are most often constructed from the rational ones (Dedekind cuts).

And the proof you're quoting shows how this basic fact implies the connectedness of certain subset of the reals, like $[0,1]$.

So you have some open (in $[0,1]$) subset $U$ in the proof (as well as some disjoint $V$ presumably), such that $0 \in U$.

You then define $A = \{t \in [0,1]: [0,t] \subseteq U \}$.

This is a non-empty subset of the reals, as $0 \in A$ (as $[0,0] = \{0\} \subseteq U$ by assumption)and bounded above (by $1$ as $A \subseteq [0,1]$).

So $\sup A $ exists. As $1$ is an upper bound and $\sup$ is the smallest upperbound by definition, $\sup A \in [0,1]$.

Now the proof goes on, deriving a contradiction (which we must do to prove connectedness) depending on whether $\sup A \in U$ or not. Both turn out to be impossible: assuming it to be in $U$, forces there to be a larger member in $A$ than $\sup A$, which cannot be, etc. See your textbook presumably.

The existence of the supremum itself is just the crucial property of the reals that is needed in the proof. You could say the existence of the supremum is what defines the reals, in a way. Doubting the validity, is doubting the validity of the real numbers themselves.

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    I think $\sup\{t\in [0,1]:[0,t]\subseteq U \}$ means the largest $t$ such that $[0,t]$ remains a subset of $U$, which is half-open on the right. So this is not the supremum of $U$, this is a number satisfying the condition that there exists a set $[0,t]$ which is the largest closed subset of $U$. But $U$ simply cannot have a largest closed subset. – 2017-01-25
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    @sequence No. this is not how it is defined. There need not be a largest such $t$, but it's an intuition about what you're trying to do in the proof. The $\sup$ exist, then a contradiction ensues. Hurrah! we have connectedness. – 2017-01-25
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    In the definition of the set, $A:=\sup\{t\in [0,1]:[0,t]\subseteq U \}$, the condition is that $[0,1]\subseteq U$. So, since $U$ is half-open on the right $\sup\{t\in [0,1]:[0,t]\subseteq U \}$ cannot be $[0, \sup(A)]$, since $\sup(A)$ is simply not included in $A$. This implies that $t$ must be less than $\sup(A)$. – 2017-01-25
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    @sequence the using of the openness is part of deriving the required contradiction. The existence of $\sup A$ is "nicht im Frage". The existence of $U$ is the problem. – 2017-01-25
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    For example, let $U=[0,1/2)$, then $\sup\{t\in [0,1]:[0,t]\subseteq U \}$ is the largest closed set $[0,t]$ such that $[0,t]\subset [0,1/2)$. Can you find such $t$? – 2017-01-25
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    @sequence No, but for $U=[0,1/2)$, $A= U$ and $\sup A=1/2$ – 2017-01-25
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Henno Brandsma's answer is excellent and addresses the question as stated. Let me address what seems to be the root of your confusion as written in the comments:

I think $\sup\{t\in[0,1]:[0,t]\subseteq U\}$ means the largest $t$ such that $[0,t]$ remains a subset of $U$

This is incorrect. Let us write $A=\{t\in[0,1]:[0,t]\subseteq U\}$. Then $\sup A$ is not necessarily an element of $A$. It is just the smallest number that is greater than or equal to every element of $A$. For instance, if $U=[0,1/2)$, then $A=[0,1/2)$ as well, and $\sup A=1/2$, which is not actually an element of $A$. So the interval $[0,\sup A]$ is not necessarily contained in $U$, and the existence of $\sup A$ does not mean there is a largest interval $[0,t]$ contained in $U$.

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    In the definition of the set, $\sup\{t\in [0,1]:[0,t]\subseteq U \}$, the condition is that $[0,1]\subseteq U$. So, since $U$ is half-open on the right $\sup\{t\in [0,1]:[0,t]\subseteq U \}$ cannot be $[0, \sup(A)]$, since $\sup(A)$ is simply not included in $A$. This implies that $t$ must be less than $\sup(A)$. – 2017-01-25
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    There is absolutely nothing wrong with $\sup A$ not being included in $A$. The definition of $\sup A$ does not require $\sup A$ to be an element of $A$. – 2017-01-25
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    I mean how can you find a $t$ such that $[0, t]$ is a closed subset of $U$, if $U$ is not closed, given that $t$ is the largest number contained in $U$? There is no largest number contained in $U$ because $U$ is not closed on the right. – 2017-01-25
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    I don't know where you got "$t$ is the largest number contained in $U$" from. What $t$ are you talking about? – 2017-01-25
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    For example, let $U=[0,1/2)$, then $\sup\{t\in [0,1]:[0,t]\subseteq U \}$ is the largest closed set $[0,t]$ such that $[0,t]\subset [0,1/2)$. Can you find such $t$? – 2017-01-25
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    As I have said repeatedly, $\sup A$ is not necessarily an element of $A$. So if $t=\sup A$, it is not necessarily true that $[0,t]\subseteq U$. – 2017-01-25
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    Correct, but the key point is that the set $\sup\{t\in [0,1]:[0,t]\subseteq U \}$ does require the "subsetness" of $U$ to be obeyed. This follows from the set-builder notation. – 2017-01-25
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    $\sup\{t\in [0,1]:[0,t]\subseteq U \}$ is not a set, it is just a number. The set $A=\{t\in [0,1]:[0,t]\subseteq U \}$ requires "subsetness of $U$", but we are not talking about $A$. We are talking about $\sup A$, which may not be an element of $A$. – 2017-01-25
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    Finally, after your most recent comment I looked at this from a different perspective and now this does indeed make sense... Thanks so much =) – 2017-01-25
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    Indeed, if one looks at the definition of $A$ as a sequence of sets, i.e. $\{[0,t]\}$ converging to $[0,t_1]$ where $t_1 < \sup(U)$ such that $\not\exists a$ satisfying $t_1$\sup(U)$ instead of $\sup\{t\in [0,1]:[0,t]\subseteq U \}$. – 2017-01-25
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    $\sup U$ might not be the same number. For instance, consider $U=[0,1/2)\cup(1/2,3/4)$. Then $\sup U=3/4$, but $\sup A=1/2$. – 2017-01-25