The final topology is finer than the initial

Question

If $T$ is a set of topologies on a (non-empty) set $X$, then the initial topology on $X$ can be defined as $\bigcup T$ while the final topology is $\bigcap T$. We see clearly that $\bigcap T \subset \bigcup T$ and thus the final topology is strictly coarser than the initial topology.

However, consider the next situation: let $X$ and $Y$ be two topological spaces with distinguished points $x_0\in X$ and $y_0\in Y$, as well as the following four maps: the ''inclusions''

$$ \begin{array}{rcl} \iota_1: X & \longrightarrow & X\times Y\\ x & \longmapsto & (x,y_0) \\ \end{array} $$ and $$ \begin{array}{rcl} \iota_2: X & \longrightarrow & X\times Y\\ y & \longmapsto & (x_0,y) \\ \end{array} $$ and the projections $$ \begin{array}{rcl} \pi_1: X\times Y & \longrightarrow & X \\ (x,y) & \longmapsto & x \\ \end{array} $$ and $$ \begin{array}{rcl} \pi_2: X\times Y & \longrightarrow & Y\\ (x,y) & \longmapsto & y . \\ \end{array} $$

We can define two topologies on $X\times Y$: the initial topology with respect to the maps $(\pi_1,\pi_2)$, with is the usual product topology and the final topology with respect to the maps $(\iota_1,\iota_2)$.

For the first, if $\mathscr B_1$ and $\mathscr B_2$ are basis of the topology of $X$ and $Y$ resp. then

$$ \mathscr B = \left\{\bigcap_{i=1}^n \pi^{-1}(U_i) : \mbox{each } U_i \mbox{ belongs to } \mathscr B_1 \mbox{ or } \mathscr B_2 \mbox{ and } n\in\mathbb N \right\} $$

is a base for such a topology.

For the second one, a bse could be

$$ \mathscr B' = \{ U\subseteq X\times Y : \iota_1^{-1}(U)\in\mathscr B_1 \mbox{ AND } \iota_2^{-1}(U)\in\mathscr B_2 \} $$

(we need the ''AND'').

What is surprising for me? That the final topology is finer than the initial. For show that let me consider an easy example: pick $X=Y=\mathbb R$ and $x_0=y_0=0$. Then, a closed set in the usual topology such as for example the square $S=[a,b]\times[a,b]$, with $0<a<b$, is open in $\mathbb R^2$ equipped with the initial topology, since $\iota_1^{-1}(S)=\iota_2^{-1}(S)=\emptyset$ which is open.

What I would like to do now? I want to show that every open set in $\mathbb R^2$ with the product topology is also open in $\mathbb R^2$ endowed with the final topology.

And I would like to do it in a smart (elegant, refined, any word sounds me good) way. For that I should consider an open set $U\subseteq\mathbb R^2$ which will be an arbitrary union of sets of the form $(a,b)\times(c,d)$ ($a<b$ and $c<d$ and may be $\infty$). And I'm not sure how to proceed here. I think my problem is that I'm not sure how to express an arbitrary open set as the union of sets belonging to $\mathscr B$.

Thanks a lot.

Answer 1

The condition on the second map is equivalent to a subset of $U$ being open if $U$ restricted to $(x_0, \mathbb{R})$ is open in $\mathbb{R}$, and the subset of $U$ restricted to $(\mathbb{R}, y_0)$ is open in $\mathbb{R}$.

In $\mathbb{R}^2$, a set is open iff it's sequentially open. So if a subset were not open in the restriction to subspaces above, then it would fail to be open in $\mathbb{R}^2$, since a sequence exhibiting non-openness would continue to exist.

Answer 2

The product topology $\mathcal{T}_p$ on $X \times Y$ is the minimal topology that makes $\pi_1$ an $\pi_2$ continuous. From this fact it follows that a function $F : Z \to X \to Y$, from any space $Z$ is continuous (in the product topology on $X \times Y$) iff $F \circ \pi_1$ and $F \circ \pi_2$ are continuous.

Now $\iota_1 \circ \pi_1 = \textrm{id}_X$ and $\iota_1 \circ \pi_2 = c_{y_0}$, where the right hand sides are the identity map and a constant map, which are both continuous, so $\iota_1$ is continuous as a map into $(X \times Y, \mathcal{T}_p)$ and similar considerations show the same for $\iota_2$.

Now the final topology $\mathcal{T}_f$ you are considering is by definition the maximal topology on $X \times Y$ that makes $\iota_1$ and $\iota_2$ continuous, and $\mathcal{T}_p$ is just some topology that does that so by maximality: $\mathcal{T}_p \subseteq \mathcal{T}_f$