Does the number $\sup\{t\in [0,1]:[0,t]\subseteq U\}$ even exist?

Question

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.

Answer 1

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 $U$.

Answer 2

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

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

Answer 3

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.