I am trying to follow a proof of the Intermediate Value Theorem in Ross's real-analysis textbook, but do not understand several of the steps. I'm going to replicate the proof as much as I can (sometimes adding additional detail, so if I state something incorrectly without knowing, please tell me) and pause at these questions.
Theorem. If $f$ is a continuous real-valued function on an interval $I$, then $f$ has the intermediate value property on $I$: Whenever $a, b \in I$, $a < b$, and $y$ lies between $f(a)$ and $f(b)$ [i.e., $f(a) < y < f(b)$ or $f(b) < y < f(a)$], there exists at least one $x$ in $(a,b)$ such that $f(x) = y$.
--
Proof. Suppose that $f(a) < y < f(b)$. We define a set $S = \{x \in [a,b]: f(x) < y\}$. Since $f(a) < y$ by assumption, $a \in S$, so $S$ is nonempty. Furthermore, $[a,b] = \{x \in \mathbb{R} : a \leq x \leq b\}$ by definition, so it follows, it follows that $x \leq b, \forall x \in S$, so $S$ is bounded above by $b$. By the completeness axiom, $S$ has a supremum, $x_0$. For any $n \in \mathbb{N}$, $x_0 - \frac{1}{n} < x_0$, so $x_0 - \frac{1}{n}$ is not an upper bound for the set of $S$, by the definition of the supremum.
The only thing I am confused on here is whether we are able to assert that $x < b$. We know $f(b) > y$, so $b$ is not in $S$, which suggests that we can make this greater assertion.
Because $x_0 - \frac{1}{n}$ is not an upper bound of $S$, there exists some element in $S$ greater than it. So, for every $n \in \mathbb{N}$, there exists some $s_n \in S$ such that $x_0 - \frac{1}{n} < s_n \leq x_0$.
Here, I am not completely sure on the strategy. He seems to take $s_n$ as not a particular element of $S$, but to treat $S$ as a sequence (we take the limit in the next step) and then take $s_n$ to be a subsequence where we create a value for each natural number. Is this the correct way to think about it?
Therefore, $\lim s_n = x_0$.
Does this follow from the squeeze lemma? Limits preserve $\leq$ inequalities, so $x_0 - \frac{1}{n} < s_n \leq x_0$ implies $\lim \left(x_0 - \frac{1}{n} \right) \leq \lim s_n \leq \lim \left(x_0 \right)$ and that $x_0 \leq \lim s_n \leq x_0$, so by the squeeze lemma $\lim s_n = x_n$.
That $f(s_n) < y$ for all $n \in \mathbb{N}$, and since $f$ is continuous, it follows that $f(x_0) = \lim f(s_n) \leq y$.
Assuming that I am correct in thinking that each $s_n$ is an element of the set that we are treating as a subsequence of $S$ (meaning that when we talk about the limit of $s_n$, we are are talking about the sequence, but when we say $f(s_n) < y$, we are talking an individual element), then I believe i understand this.
Define $t_n = \min\left(b, x_0 + \frac{1}{n}\right)$.
I am assuming, again, that this is a subsequence of $S$.
Observe that $x_0 \leq t_n \leq x_0 + \frac{1}{n}$.
I do not understand completely the second of these inequalities. If $t_n = b$, since $b$ is an upper bound of $S$ and $x_0$ is its supremum, it follows that $x_0 \leq t_n$, so I understand this. But $n$ is strictly positive, obviously, so $x_0 + \frac{1}{n} > x_0$. If $x_0 = t_n$, then $x_0 + \frac{1}{n} > t_n$. I suppose we have $t_n = x_0 + \frac{1}{n}$ is $x_0 + \frac{1}{n} < b$ and thus $t_n = x_0 + \frac{1}{n}$. Is this correct?
Therefore, $\lim t_n = x_0$.
I assume this is also a result of the squeeze lemma, correct?
Observe that each $t_n$ belongs to $[a,b]$ but not to $S$.
If $t_n = b$, then it surely doesn't belong to $S$ since $f(b) > y$. Is the additional argument then that since $x_0$ is the supremum of $S$, then if $x_0 + \frac{1}{n}$ were in $S$, then $x_0$ couldn't be its supremum (a contradiction), then $x_0 + \frac{1}{n}$ cannot be in the set? I think I understand why $t_n$ is not in $S$, but not why they are still in $[a,b]$. Surely $b$ is in $[a,b]$, and I suppose in the case where $x_0 + \frac{1}{n}$ is greater than $b$, then $t_n = b$, so we stay in $[a,b]$. Is that correct?
Therefore, $f(t_n) \geq y$ for all $n$. Therefore, since $f$ is continuous, $f(x_0) = \lim f(t_n) \geq y$. Since $f(x_0) \leq y$ and $f(x_0) \geq y$, it follows that $f(x_0) = y$. Therefore, $\exists x \in (a,b)$ such that $f(x) = y$.
I understand everything here.
I would greatly appreciate any insights on the above points or on something I missed. Thanks in advance.
Here are my comments on your arguments:
If $b$ were in the set $S$, then it would be that $f(b)<y$ which contradicts our assumption. So yes, your conclusion is true.
No, we are trying to construct a sequence which converges to $x_0$. How do we do it? Since for each $n \in \mathbb{N}$, $x_0 - 1/n$ is no longer an upperbound for $S$, we can choose $s_n \in S$ such that $x_0 - 1/n < s_n \le x_0$. Can you see that we constructed a sequence $(s_n)$? And btw, $S$ is a set, it isn't a sequence.
You are right. But there's a typo at the end of previous paragraph: $\lim s_n = x_0$.
Yes, $s_n \in S$ for all $n \in N$, so by definition of $S$, we have $f(s_n) < y$ for all $n \in \mathbb{N}$. To conclude that $f(x_0) = \lim f(s_n) \leq y$, we first notice that $f(s_n) < y$ for all $n \in \mathbb{N}$ then it must be that $\lim f(s_n) \le \lim y = y$ (since $f(s_n)$ is a sequence too!) and by the continuity of $f$, we have $f(s_n) = f(x_0)$
Yet again, $S$ is a set and $(t_n)$ is a sequence.
No, you are incorrect here. Let's first discuss about the first inequality. Since $x_0$ is the lub for $S$ and $b$ is an upperbound for $S$, we have that $x_0 \le b$. Since $x_0 \le b$ and $x_0 < x_0 + 1/n$ for all $n \in \mathbb{N}$, it must be that $x_0 \le \min \{ b, x_0 + 1/n \} = t_n$. The second inequality follows immediately from the definition of minimum, $t_n = \min \{ b, x_0 + 1/n \} \le x_0 + 1/n$.
Yes, you are correct if $x_0 + 1/n$ was in $S$, then $x_0$ would no longer be an upperbound for $S$. The sequence $(t_n)$ is in the set $[a,b]$ because $a \le x_0 \le t_n = \min \{ b, x_0 + 1/n \} \le b$.