Alternative proof for subbasis for order topology and product topology: Finite intersections of elements of $\mathscr S$ is a basis

Question

Topology by James Munkres:

Both of the following proofs are at the very end of, respectively, Sections 14 and 15, which seem to correspond with the definition of a subbasis' being at the very end of Section 13.

1. Open rays are a subbasis for the order topology

2. Thm 15.2 The following collection $\mathscr S$ is a subbasis for the product topology.

$$\mathscr S = \{\pi_{1}^{-1}(U):\text{U open in X}\}\cup\{\pi_{1}^{-1}(V):\text{V open in Y}\} \tag{*}$$

It seems like the proofs are showing that $\mathscr T = \mathscr T'$ by showing $\mathscr T \subseteq \mathscr T'$ and $\mathscr T \supseteq \mathscr T'$, where $\mathscr T$ is the order topology or the product topology while $\mathscr T'$ is, respectively, the topology generated by the open rays or the topology generated by the $\mathscr S$ in $(*)$, by making use of Lemma 13.3 as follows.

Can we alternatively show that the collection $\mathscr B$ of all finite intersections of elements of $\mathscr S$ is a basis for $\mathscr T$, by making use of Lemma 13.2 as follows? Or are the methods actually equivalent?

My idea is based on, well, the very end of Section 13.

Answer 1

Proof that open rays are a subbasis $\mathscr S$ for the order topology $\mathscr T$ by showing that $\mathscr B$ of all finite intersections of elements of $\mathscr S$ is a basis for $\mathscr T$:

(Hopefully, it is obvious that $\mathscr S$ is a subbasis for some topology on $X$.)

First, $\mathscr B = \{\text{open rays in X}, \text{open intervals in X}\}$, where some (resp: all) open rays are half-open intervals if and only if $X$ has a $\max$ or (resp: and) $\min$. Also, denote $\mathscr B_{\text{ord}}$ to be the basis that defines $\mathscr T$.

To show $\mathscr B$ is a basis for the order topology $\mathscr T$, we must show that for all $A \in \mathscr T$ and $x \in A$, we can find $B \in \mathscr B$ s.t. $x \in B \subseteq A$.

Let $A \in \mathscr T$ and $x \in A$. By def of $A$'s being an element of $\mathscr T$, we have that for all $x \in A$, we can find $B_{\text{ord}} \in \mathscr B_{\text{ord}}$ s.t. $x \in B_{\text{ord}} \subseteq A$. Now, we wonder how $B_{\text{ord}}$ can help us find $B$. Well, by def of $B_{\text{ord}}$'s being an element of $\mathscr B_{\text{ord}}$, such $B_{\text{ord}}$ is in one of the following three forms:

$1. B_{\text{ord}} = (a,b)$

Obviously, choose $B=B_{\text{ord}}$.

$2. B_{\text{ord}} = [a_0,b)$

This expression makes sense if and only if $X$ has a minimum given by $a_0 := \min X$. Thus, by def of the open ray $(-\infty,b)$, we have that $[a_0,b)=(-\infty,b)$. Therefore, we can choose, but not quite obviously, $B=B_{\text{ord}}$.

$3. B_{\text{ord}} = (a,b_0]$

Similar to (2):

This expression makes sense if and only if $X$ has a maximum given by $b_0 := \max X$. Thus, by def of the open ray $(a,\infty)$, we have that $(a,b_0]=(a,\infty)$. Therefore, we can choose, but not quite obviously, $B=B_{\text{ord}}$.

Therefore, $\mathscr B$ is a basis for the order topology $\mathscr T$, by Lemma 13.2 (which I'm not quite sure is different from the definition of a basis itself, but whatever).

Therefore, $\mathscr S$ is a subbasis for $\mathscr T$.

QED

Considering that this proof relies on the same observations (the three forms) as the original proof, the proofs look pretty equivalent to me.

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Proof that the following collection $\mathscr S$ is a subbasis for the product topology $\mathscr T$

$$\mathscr S = \{\pi_{1}^{-1}(U):\text{U open in X}\}\cup\{\pi_{1}^{-1}(V):\text{V open in Y}\} \tag{*}$$

by showing that $\mathscr B$ of all finite intersections of elements of $\mathscr S$ is a basis for $\mathscr T$:

(Hopefully, it is obvious that $\mathscr S$ in (*) is a subbasis for some topology on $X \times Y$.)

First, $\mathscr B = \{ \bigcap_{i_j=i_1}^{i_n} S_{i_j} : S_{ij} \in \mathscr S \}$. Also, denote $\mathscr B_{X \times Y}$ to be the basis that defines $\mathscr T$ and $\mathscr T_X$ and $\mathscr T_Y$ as any topologies of, resp, $X$ and $Y$.

To show $\mathscr B$ is a basis for the product topology $\mathscr T$, we must show that for all $A \in \mathscr T$ and $(x,y) \in A$, we can find $B \in \mathscr B$ s.t. $(x,y) \in B \subseteq A$.

Let $A \in \mathscr T$ and $(x,y) \in A$. By def of $A$'s being an element of $\mathscr T$, we have that for all $(x,y) \in A$, we can find $B_{X \times Y} \in \mathscr B_{X \times Y}$ s.t. $(x,y) \in B_{X \times Y} \subseteq A$. Now, we wonder how $B_{X \times Y}$ can help us find $B$. Well, by def of $B_{X \times Y}$'s being an element of $\mathscr B_{X \times Y}$, there exist $U_B \in \mathscr T_X, V_B \in \mathscr T_Y$ s.t. $B_{X \times Y}$ is given by

$$B_{X \times Y} = \pi_{1}^{-1}(U_B) \cap \pi_{2}^{-1}(V_B) \tag{**},$$

where

$\pi_{1}^{-1}(U_B) \in \mathscr S$ for $V=\emptyset=\pi_{1}^{-1}(\emptyset)=X \times \emptyset$ and $\pi_{2}^{-1}(V_B) \in \mathscr S$ for $U=\emptyset=\pi_{2}^{-1}(\emptyset)=\emptyset \times Y$

Therefore, $\mathscr B$ is a basis for the product topology $\mathscr T$, by Lemma 13.2 (which I'm not quite sure is different from the definition of a basis itself, but whatever).

Therefore, $\mathscr S$ in (*) is a subbasis for $\mathscr T$.

QED

Considering that this proof relies on the same observation $(**)$ as the original proof, the proofs look pretty equivalent to me.

Answer 2

In general if $\tau$ denotes a topology and $\mathcal B$ denotes a basis of $\tau$ then - if you want to prove that $\mathcal S$ is a subbasis of $\tau$ and $\mathcal V$ denotes the collection of finite intersections of elements of $\mathcal S$ - it is enough to prove that every $B\in\mathcal B$ can be written as union of elements of $\mathcal V$.

This because then it is guaranteed that also that every $O\in\tau$ (that by definition can be written as union of basis elements) can be written as union of elements of $\mathcal V$.

We get something like:$$O=\bigcup_{i\in I}B_i=\bigcup_{i\in I}\bigcup_{j\in J_i}V_{i,j}$$where the $B_i\in\mathcal B$ and the $V_{i,j}\in\mathcal V$.

In your situation we have $\mathcal B=\mathscr B_{\text{ord}}\subseteq\mathscr B=\mathcal V$ showing directly that every element of $\mathcal B$ can be written as union of elements of $\mathcal V$.