I am reading "Topology 2nd Edition" by James R. Munkres.
I read the proof of Theorem 16.4 on p.91 and the proof was difficult for me.
So, I filled very small gaps in the proof and rewrote the proof.
Is my proof ok?
Theorem 16.4. $\,\,\,\,\,$Let $X$ be an ordered set in the order topology; let $Y$ be a subset of $X$ that is convex in $X$. Then the order topology on $Y$ is the same as the topology $Y$ inherits as a subspace of $X$.
Proof.$\,\,\,\,\,$Consider the ray $(a,+\infty)$ in $X$. What is its intersection with $Y$? If $a\in Y$, then $$(a,+\infty)\cap Y=\{x\mid x\in Y \text{ and }x>a\};$$ this is an open ray of the ordered set $Y$. If $a\notin Y$, then $a$ is either a lower bound on $Y$ or an upper bound on $Y$, since $Y$ is convex. In the former case, the set $(a,+\infty)\cap Y$ equals all of $Y$; in the latter case, it is empty.
$\,\,\,\,\,$A similar remark shows that the intersection of the ray $(-\infty,a)$ with $Y$ is either an open ray of $Y$, or $Y$ itself, or empty. Since the sets $(a,+\infty)\cap Y$ and $(-\infty,a)\cap Y$ form a subbasis for the subspace topology on $Y$, and since each is open in the order topology, the order topology contains the subspace topology.
$\,\,\,\,\,$To prove the reverse, note that any open ray of $Y$ equals the intersection of an open ray of $X$ with $Y$, so it is open in the subspace topology on $Y$. Since the open rays of $Y$ are a subbasis for the order topology on $Y$, this topology is contained in the subspace topology.
I filled small gaps in the above proof:
My proof:
Proposition 1:
The sets $(a,+\infty)\cap Y$ and $(-\infty,a)\cap Y$ form a subbasis for the subspace topology on $Y$, where $(a,+\infty)$ and $(-\infty,a)$ are open rays in $X$.
Proof:
Let $y$ be an arbitrary element of $Y$.
Since $X$ has more than one element by the definition of the order topology on p.84, there exists $a\in X$ such $a<y$ or $y<a$.
So, $y\in (a,+\infty)$ or $y\in (-\infty,a)$ for some $a\in X$.
So, $y\in (a,+\infty)\cap Y$ or $y\in (-\infty,a)\cap Y$ for some $a\in X$.
So, the sets $(a,+\infty)\cap Y$ and $(-\infty,a)\cap Y$ form a subbasis for a topology $\mathcal{T}$ on $Y$.
We prove $\mathcal{T}$ equals the subspace topology on $Y$.
By Lemma 16.1 on p.89, the sets $(a,b)\cap Y$ and $[a_0,b)\cap Y$ and $(a,b_0]\cap Y$ form a basis for the subspace topology on $Y$, where $(a,b)$ and $[a_0,b)$ and $(a,b_0]$ are intervals in $X$ and $a_0$ is the smallest element (if any) of $X$ and $b_0$ is the largest element (if any) of $X$.
Since $(a,b)\cap Y=((a,+\infty)\cap Y)\cap ((-\infty,b)\cap Y)$ and $[a_0,b)\cap Y=(-\infty,b)\cap Y$ and $(a,b_0]\cap Y=(a,+\infty)\cap Y$, any element of the subspace topology on $Y$ is an element of $\mathcal{T}$.
Since $(a,+\infty)$ and $(-\infty,a)$ are open sets of $X$ (See p.86), $(a,+\infty)\cap Y$ and $(-\infty,a)\cap Y$ are elements of the subspace topology on $Y$.
So, any element of $\mathcal{T}$ is an element of the subspace topology on $Y$.
Proposition 2:
$(a,+\infty)\cap Y$ and $(-\infty,a)\cap Y$ are open in the order topology on $Y$.
Proof:
For example, let $X:=\mathbb{R}$ and $Y:=[0,1]\cup [2,3]$.
Then, $(\frac{3}{2},+\infty)\cap Y=[2,3]$ is not open in the order topology on $Y$.
So, we need to prove the proposition 2.
If $a\in Y$, then $(a,+\infty)\cap Y=\{x\mid x\in Y \text{ and }x>a\}$ is an open ray of the orderd set $Y$.
So, $(a,+\infty)\cap Y$ is open in the order topology on $Y$ (See p.86).
If $a\notin Y$, then $a$ is either a lower bound on $Y$ or an upper bound on $Y$.
We prove this.
Assume that $a$ is not a lower bound on $Y$ and $a$ is not an upper bound on $Y$.
Then, there exist $y_1,y_2\in Y$ such that $y_1<a<y_2$.
Since $Y$ is convex in $X$, $a\in Y$.
This is a contradiction.
If $a$ is a lower bound on $Y$, then $(a,+\infty)\cap Y=Y$.
So, in this case $(a,+\infty)\cap Y$ is open in the order topology on $Y$.
If $a$ is an upper bound on $Y$, then $(a,+\infty)\cap Y=\emptyset$.
So, in this case $(a,+\infty)\cap Y$ is open in the order topology on $Y$.
Similarly we can show that $(-\infty,a)\cap Y$ is open in the order topology on $Y$.
Proposition 3:
Any open ray of $Y$ is open in the subspace topology on $Y$.
Proof:
We can write any open ray of $Y$ as $(a,+\infty)\cap Y$ or $(-\infty,a)\cap Y$ for some $a\in Y$.
Since $(a,+\infty)$ and $(-\infty,a)$ are open in $X$ (See p.86), any open ray of $Y$ is open in the subspace topology on $Y$.
Proposition 4:
The open rays of $Y$ are a subbasis for the order topology on $Y$.
Proof:
See p.86.
From proposition 1 and proposition 2, it is obvious that the order topology on $Y$ contains the subspace topology on $Y$.
From proposition 3 and proposition 4, it is obvious that the subspace topology on $Y$ contains the order topology on $Y$.
