Some Questions from the proof of the result : The unit interval $\mathbb I = [0,1]$ is compact The unit interval $\mathbb I = [0,1]$ is compact
I was trying to understand the proof of the above result from my textbook which goes like as follows. However, I have a few questions in mind. Please guide me on the questions shown in blue :
Proof :Let ${\cal C}$ be an open cover of $[0,1]$.
Let $T =\sup \{ t \in [0,1]| [0,t] \text{ can be covered by a finite subset of }{\cal C} \} $.
Since $0 \in U$ for some $U \in {\cal C}$, we see that $T > 0$.
Suppose $T < 1$. Then $T \in U'$ for some $U' \in {\cal C}$, and hence :
$(T-\epsilon, T+\epsilon) \subset U'$ for some $\epsilon >0$.
However, this contradicts the definition of $T$. Hence $T = 1$.

$(i)~~$How can we be sure that there will surely exist a $t  \in [0,1]$ such that $[0,t]$ can be covered by a finite subset of $ \cal C $ ? And then we named the supremum of such a $t \in [0,1]$ as $T$.
$(ii)~~$Even if it's possible to express an interval $[0,t]$ as the finite union of open intervals, then shouldn't the value of $T=1$ as $[0,1] \subset (-1,1.5)?$
$(iii)~~$ We said that suppose $T < 1$. Then $T \in U'$ for some $U' \in {\cal C}$, and hence $(T-\epsilon, T+\epsilon) \subset U'$ for some $\epsilon >0$
Then does this contradict the definition of $T$ because the defined supremum then becomes $T + \epsilon$?

Thank you for your help.
 A: $(i)$ Since $0$ is in one of the open sets, there is an open neighborhood $[0,\epsilon)$ that is contained in that open set. Any $t$ in that interval is in the set in question, so such a $t$ exists. Since $[0,1]$ is bounded above, the set of such $t$ has a least upper bound.
$(ii)$ There is no a priori reason to believe that $t=1$ is in $T$; that is, that $[0,1]$ can be covered by a finite number of $\mathcal{C}$. That is what you are trying to prove. Or perhaps I don't understand what your question is.
$(iii)$ Yes, that's exactly the point. You assume $T$ is the supremum, and then show that some number greater than $T$ is in the set. This contradicts the assumption that $T<1$ and it follows that $T=1$, so that $[0,1]$ can be covered by only a finite number of elements of $\mathcal{C}$.
A: (i) It $t=0$ then the interval $[0,0]$ is just a the point $0$ which is in the interval. Since $C$ is an open cover, there must be at least one element in $C$ that contains $0$. This means that the set $\{t\in[0,1]|[0,t] \text{ can be covered by a finite subset of }C\}$ is non empty. A nonempty subset of a bounded set has a supremum, so $T$ exists.
(ii) I think your issue in this part had to do with the fact that $C$ is an arbitrary open cover. That is $(−1,1.5)$ may not be in $C$, so a priori, you do not know what $T$ is. All you know is that it exists.
(iii) The last two lines of the proof: "Suppose $T<1$. Then $T$..." could be preceded by "Suppose (for the sake of contradiction) that $T<1$". Then yes, you get a contradiction saying that $T+\epsilon$ is actually the supremum. That is, you can contradict the statement "$T<1$", and thus $T$ is NOT stricly less than 1. Since $T$ is in $[0,1]$, then that means that $T$ is what?
A: *

*We know that $[0,0]$ can be covered by a finite subset of $\mathcal C$, namely by any open ball containing $0$. So we know there exists $t$ so that $[0,t]$ is covered by a finite subset of $\mathcal C$.

*I don't understand your complaint. Yes, $[0,t]$ can be covered by a finite union of open balls in $\mathcal C$. The goal of the proof is to show that $T=1$. I don't see how this has anything to do with $(-1, 1.5)$.

*Yes. $T < T + \epsilon$, contradiction.
