On the proof of "A subset of $\mathbb R$ is pathwise-connected iff is an interval."

Question

From the book Advanced Calculus by Patrick Fitzpatrick :

Theorem:

A subset of $\mathbb R$ is pathwise-connected if and only if it is an interval.

Proof:

First, suppose that $I$ is an interval. Let $u$ and $v$ be points in $I$. We can suppose that $u < V$. Since $I$ is an interval, the closed interval $[u,\ v]$ is a subset of $I$, so we can define $[u,v]$ to be the parameter space and define $\gamma (t) = t$ for $u \le t \le v$ to obtain a path in $I$ joining the points $u$ and $v$.

To prove the converse, suppose that $I$ is a pathwise-connected subset of $\mathbb R$. To verify that $I$ is an interval, we select two points $u$ and $v$ in $I$ with $u < v$. We must show that the closed interval $[u,v]$ is a subset of $I$. Since $I$ is pathwise-connected, there is a parametrized path $\gamma[a,b] \rightarrow I$ with $\gamma (a) = u$ and $\gamma (b) = v$. Now the Intermediate Value Theorem asserts that $[ \gamma (a), \gamma (b)] \subseteq \gamma [a,b]$ so $[u,v] \subseteq I$.

I don't understand the last sentence at all. How Intermediate Value Theorem results in $[ \gamma (a), \gamma (b)] \subseteq \gamma [a,b]$? How this is derived?

Answer 1

The intermediate value theorem asserts that $[\gamma(a),\gamma(b)] \subseteq \gamma[a,b]$ for any continuous $\gamma$ and $a,b$ in the domain of $\gamma$. It is derived here.

Answer 2

The intermediate value theorem says for f continuous: $$u\in[f(a),f(b)]\implies \exists x\in[a,b]:f(x)=u.$$ Thus we have $[f(a),f(b)]\subseteq f([a,b])$. In your case we obtain $$[u,v] = [\gamma(a), \gamma(b)] \subseteq \gamma([a,b]) \subseteq I.$$