If $a\leq b \leq c$, and $c-a$ can be made arbitrarily small does that mean $a=b=c$? What is the actual thing that justifies this inference? The question is pretty clear I think but for context, I'm reading chapter 13 of Spivak's Calculus, entitled "Integrals" and came across a seemingly simple step in a proof which left me wondering about why it was justified.
A function $f$ that is bounded on an interval $[a,b]$ is defined as integrable on that interval if
$$\sup\{L(f,P)\}=\inf\{U(f,P)\}\tag{1}$$
where $P$ denotes any partition of $[a,b]$, $L(f,P)$ denotes the lower sum of $f$ for partition $P$, and $U(f,P)$ denotes the upper sum of $f$ for partition $P$.
The number represented by $(1)$ is the integral of $f$ on $[a,b]$, and is denoted $\int_a^b f$.
There is also a theorem that in essence restates the definition

Theorem 2: if $f$ is bounded on $[a,b]$ then, $f$ is integrable on
$[a,b]$ if and only if for any $\epsilon>0$ there exists a partition
$p$ such that $U(f,P)-L(f,P)<\epsilon$

At some point, Spivak presents the following theorem

Theorem 5: If $f$ and $g$ are integrable on $[a,b]$, then $f+g$ is
integrable on $[a,b]$ and
$$\int_a^b (f+g)=\int_a^b f + \int_a^b g$$

I have a question about a seemingly very simple step at the end of the proof.
Here is the proof of the first part, that $f+g$ is integrable.

Since $f$ and $g$ are integrable, there are partitions $P'$ and $P''$
such that
$$U(f,P')-L(f,P')<\frac{\epsilon}{2}$$
$$U(g,P'')-L(g,P'')<\frac{\epsilon}{2}$$
Let $Q$ be the partition containing $P'$ and $P''$. Then from a lemma
presented in the text, we can infer that
$$U(f,Q) \leq U(f,P')$$ $$L(f,Q) \geq L(f,P')$$
and
$$U(g,Q) \leq U(g,P'')$$ $$L(g,Q) \geq L(g,P'')$$
Hence,
$$U(f,Q)+U(g,Q)\leq U(f,P')+U(g,P'')$$ $$L(f,Q)+L(g,Q) \geq
 L(f,P')+L(g,P'')$$
$$U(f,Q)+U(g,Q)-(L(f,Q)+L(g,Q)) \leq
 U(f,P')+U(g,P'')-(L(f,P')+L(g,P''))<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$
That is, $U(f+g,Q)-L(f+g,Q)<\epsilon$. So $f+g$ is integrable on
$[a,b]$.

Here is the proof of the second part, that $\int_a^b (f+g)=\int_a^b f + \int_a^b g$.

Let $P={t_0,...,t_n}$ be any partition of $[a,b]$.
Let
$$m_i=\inf\{ (f+g)(x): t_{i-1} \leq x \leq t_i \}$$ $$m_i'=\inf\{
 (f)(x): t_{i-1} \leq x \leq t_i \}$$ $$m_i''=\inf\{ (g)(x): t_{i-1}
 \leq x \leq t_i \}$$
$$M_i=\sup\{ (f+g)(x): t_{i-1} \leq x \leq t_i \}$$ $$M_i'=\sup\{
 (f)(x): t_{i-1} \leq x \leq t_i \}$$ $$M_i''=\sup\{ (g)(x): t_{i-1}
 \leq x \leq t_i \}$$
It can be shown that $m_i \geq m_i'+m_i''$ and $M_i\leq M_i'+M_i''$.
Therefore,
$$L(f,P)+L(g,P) \leq L(f+g,P)\tag{2}$$
$$U(f+g,P) \leq U(f,P)+U(g,P)\tag{3}$$
Putting $(1)$ and $(2)$ together we have
$$L(f,P)+L(g,P) \leq L(f+g,P) \leq U(f+g,P) \leq U(f,P)+U(g,P)$$
Now, by definition of the integral $\int_a^b f$ and $\int_a^b g$ we
have
$$L(f,P)\leq \int_a^b f \leq U(f,P)$$
$$L(g,P)\leq \int_a^b g \leq U(g,P)$$
Therefore,
$$L(f+g,P)=L(f,P)+L(g,P)\leq \int_a^b f +\int_a^b g \leq U(f,P)
 +U(g,P)=U(f+g,P)\tag{4}$$
But since $f+g$ is integrable, then for (any) P we have
$$L(f+g,P) \leq \int_a^b (f+g) \leq U(f+g,P)\tag{5}$$

At this point, intuition would tell us that $\int_a^b (f+g)=\int_a^b f +\int_a^b g$.
Here is how Spivak justifies this last step:

Since $U(f,P)-L(f,P)$ and $U(g,P)-L(g,P)$ can both be made as small as
desired, it follows that
$$U(f,P)+U(g,P)-[L(f,P)+L(g,P)]$$
can also be made as small as desired; it therefore follows from $(4)$
and $(5)$ that
$$\int_a^b (f+g)=\int_a^b f +\int_a^b g$$

I can sort of see intuitively that if $a\leq b \leq c$, and $c-a$ can be made arbitrarily small then that should mean that $a=b=c$. But what is the actual thing that justifies this inference?
 A: Speaking to this in the title and at the end (I haven't read the whole question)

I can sort of see intuitively that if $≤≤$ , and $−$ can be made
arbitrarily small then that should mean that $==$. But what is the
actual thing that justifies this inference?

Ignore $b$ for the moment. If the difference between $a$ and $c$ is less than $\epsilon$ for every $\epsilon$ then they must be equal. If not, take $\epsilon = (c-a)/2$ to reach a contradiction.
Since $b$ is between $a$ and $c$ and $a=c$, $b$ must be equal to $a$ as well.
A: 
if $a\leq b \leq c$, and $c-a$ can be made arbitrarily small then that should mean that $a=b=c$.

Precisely, this means

if $a\leq b \leq c$, and $c-a<\varepsilon$ for every $\varepsilon>0$, then $a=b=c$.

The trick is that, if $a\neq c$, then $a<c$. Consider $\varepsilon=\frac{c-a}{2}>0$. Then $c-a<\varepsilon \iff c-a<\frac{c-a}{2}$ which is absurd since $c-a>0$.
Therefore, $c=a$. Since $a\le b\le c$, anti-symmetry of the order relation implies that $a=b=c$.
A: We start with the following statements
$$L(f+g,P)\leq \int_a^b f +\int_a^b g \leq U(f+g,P)\tag{4}$$
$$L(f+g,P) \leq \int_a^b (f+g) \leq U(f+g,P)\tag{5}$$
where $(4)$ and $(5)$ are true for any partition $P$.
and the fact that $f+g$ is integrable, which means that $\forall \epsilon>0$, we can find some partition $P$ such that
$$U(f,P)-L(f,P)<\epsilon\tag{6}$$
This situation can be seen in a simplified manner by letting
$$U(f+g,P)=c$$
$$L(f+g,P)=a$$
$$\int_a^b (f+g)=b_1$$
$$\int_a^b f +\int_a^b g=b_2$$
Then we have
$$a \leq b_1\leq c\tag{1}$$
$$a \leq b_2\leq c\tag{2}$$
and $c-a$ can be made arbitrarily small.
Assume $b_1 \neq b_2$.
Case 1: $b_1<b_2$
Let $d \in (b_1,b_2)$.
Then if we choose $\epsilon=\frac{b_2-b_1}{2}$ we have
$$c-a<\frac{b_2-b_1}{2}$$
$$c<a+\frac{b_2-b_1}{2}<b_1+\frac{b_2-b_1}{2}=\frac{b_1+b_2}{2}<b_2$$
$$\bot$$
Case 2, $b_2<b_1$ is symmetric to case 1, and thus results in $\bot$ as well.
Thus, by contradiction, we conclude that $b_1=b_2$.
Finally note the following distinction.
If we are given $a$ and $c$ and we say that for these specific numbers it is true that $c-a<\epsilon$ for any $\epsilon$, then as shown by Taladris's answer, it must be the case that $a=c$. If there is some $b$ such that $a \leq b \leq c$ then $a=b=c$.
There is a slightly different setup: a given $b$ and variable $a$ and $c$ such that $a \leq b \leq c$ and it is always possible to choose $c$ and $a$ such that $c-a<\epsilon$ for any $\epsilon$. In this case, $a$ doesn't have to be equal to $c$.
For example, let $a(n)=b(1-\frac{1}{n})$ and $c(n)=b(1+\frac{1}{n})$.
We can make $a(n)$ and $c(n)$ as close to each other as we want but $a(n)<c(n)$ for all $n$.
What we can show for this second case, however, is what I showed above: that there can be at most one number between $a(n)$ and $c(n)$.
