Is this proof for Theorem 16.4 Munkers Topology correct?

The followings is the Theorem 16.4 from Munkers’ Topology:

enter image description here

In the textbook it uses concept of subbasis to prove the theorem which I can’t understand it. I tried to prove that in another way but I am not sure if I am correct.

Proof?:

1- Both $Y$ and $\emptyset$ are open in both typologies.

2- Basis elements in order topology in $Y$ are one of the three types: (1) All open intervals $(a,b)$ in $Y$; (2) All intervals of the form $[a_0,b)$, where $a_0$ is the smallest element (if any) of Y; (2) All intervals of the form $(a,b_0]$, where $b_0$ is the largest element (if any) of Y. Each of them can be written as $(c,d) \cap Y$ for $(c,d) \subset X$ an open set in X. And since $(c,d) \cap Y$ is open so the topology Y inherited as a subspace of X is finer than order topology in $Y$.

3- Any open set the topology Y inherited as a subspace of X is in the form $(c,d) \cap Y$ for $(c,d) \subset X$ an open set in X. This intersection is either Y or $\emptyset$ or one the mentioned types of basis elements in order topology in $Y$ in paragraph 2. So the topology Y inherited as a subspace of X is coarser than order topology in $Y$. And we are done!

Is this proof a correct and rigorous one? And if not, is there any other simple proof not using subbasis?

Solutions Collecting From Web of "Is this proof for Theorem 16.4 Munkers Topology correct?"

You could stand to give a bit more detail. Let $\tau_Y$ be the subspace topology on $Y$ and $\tau$ the order topology on $Y$, and for $a,b\in Y$ write $(a,b)_Y$, $(a,b]_Y$, etc. for intervals in the relative order on $Y$. In showing that $\tau\subseteq\tau_Y$, you can specify precisely how the intervals in $Y$ are obtained from intervals in $X$.

  • If $a,b\in Y$ with $a<b$, then $(a,b)_Y=(a,b)\cap Y\in\tau_Y$.
  • If $a_0=\min Y$, and $b\in Y\setminus\{a_0\}$, then either $a_0=\min X$, in which case $[a_0,b)_Y=[a_0,b)\cap Y\in\tau_Y$, or there is an $a\in(\leftarrow,a_0)$, in which case $[a_0,b)_Y=(a,b)\cap Y\in\tau_Y$.
  • The case in which $b_0=\max Y$, $a\in Y\setminus\{b_0\}$, and we consider $(a,b_0]_Y$ is similar.

For the proof that $\tau_Y\subseteq\tau$, note that it is not true that every open set in $Y$ is of the form $(c,d)\cap Y$ for some $c,d\in X$ with $x<d$. It isn’t even quite true that these sets form a base for $\tau_Y$: that’s the case if and only if $X$ has neither a minimum nor a maximum element. It’s fine to work with a base for $\tau_Y$, of course, but you do have to split it into three cases, just as in the other part of the proof. Moreover, since $c$ and $d$ need not be in $Y$, even showing that $(c,d)\cap Y\in\tau$ takes a bit of work that you’ve not supplied.

Let $c,d\in X$ with $c<d$, and let $B=(c,d)\cap Y\in\tau_Y$. If $c,d\in Y$, then the convexity of $Y$ ensures that $B=(c,d)_Y\in\tau$, but there’s no reason to assume that $c$ and $d$ are in $Y$. To show that $B\in\tau$, we’ll show that for each $y\in B$ there is a $U_y\in\tau$ such that $y\in U_y\subseteq B$; $B$ will then be a union of members of $\tau$ and hence itself in $\tau$. Let $y\in B$; clearly $c<y<d$.

  • If $y$ is neither the minimum nor the maximum element of $B$, there are $u,v\in B$ such that $c<u<y<v<d$. $Y$ is convex, and $u,v\in Y$, so $(u,v)\subseteq Y$, and therefore $(u,v)_Y=(u,v)$. Thus, $y\in(u,v)_Y\subseteq B$, where certainly $(u,v)_Y\in\tau$.

  • If $y=\min B$, and there is some $v\in B$ such that $y<v$, there are two possibilities. One is that $y=\min Y$; in that case $[y,v)\subseteq Y$, so $y\in[y,v)_Y\subseteq B$, and $[y,v)_Y\in\tau$. The other is that $y\ne\min Y$, in which case there is some $u\in Y$ such that $u<y$. Clearly $u\le c$, since $y=\min B$, and the convexity of $Y$ then implies that $c\in Y$. But then $y\in(c,v)_Y\subseteq B$, and $(c,v)_Y\in\tau$. (It’s not hard to show that if $y\ne\min Y$, then in fact not only is $c\in Y$, but $c$ is the immediate predecessor of $y$ in $X$: the open interval $(c,y)=(c,y)_Y=\varnothing$.)

  • If $y=\max B$, and there is some $u\in B$ such that $u<y$, you can show similarly that either $y=\max Y$, and $y\in(u,y]_Y\subseteq B$ with $(u,v]_Y\in\tau$, or $d\in Y$, and $y\in(u,d)_Y\subseteq B$ with $(u,d)_Y\in\tau$.

  • The only remaining possibility is that $y=\min B=\max B$, so that $B=\{y\}$. In this case you can use the same ideas to show that $c,d\in Y$, so that $B=(c,d)_Y\in\tau$.

It really is easier to work with the subbase of open rays: there are fewer cases that need to be handled separately.