Question about infinite product topology

Question

The product topology, sometimes called the Tychonoff topology, on ${\textstyle \prod _{i\in I}X_{i}}$ is defined to be the coarsest topology (that is, the topology with the fewest open sets) for which all the projections ${\textstyle p_{i}:\prod X_{\bullet }\to X_{i}}$ are continuous.(https://en.wikipedia.org/wiki/Product_topology) We can show that there exists topology satisfying the 'continue' conditions for all projection(discrete topology).

I see some proof using $\beta=\{ {\textstyle \prod _{i\in I}U_{i}}; U_i \in \tau_i\}$ where each $U_{i}$ is open in $X_{i}$ and ${\displaystyle U_{i}\neq X_{i}}$ for only finitely many $i$. My question is

I can show that $\beta$ is a topology basis and the generated topology satisfying the 'continue' conditions. How can I show that it is the coarsest topology.

I go as follows, suppose we have another topology satisfies the 'continue' condition, namely $\tau$, now I need to show $\tau_{\beta} \subset\tau$.

$\tau$ contains open sets $\prod _{i\neq j}X_{i} \times U_j$ Since all open sets in $\tau_{\beta}$ can be expressed as $\cup \prod _{i\in I}U_{i}$, then I have no ides.

Answer 1

The coarsest topology exists because any intersection of topologies on a fixed set $X$ is again a topology on $X$. The coarsest topology for which the projections are continuous is simply the intersection of all the topologies for which the projections are continuous.

Since projections are continuous with respect to $\tau$, we find that $\pi^{-1}(U_{i})=U_{i}\times\prod_{j\neq i}X_{j}$ is in $\tau$ for any $U_{i}\in\tau_{i}$. Therefore, any finite intersection of such sets is also in $\tau$. Such a finite intersection looks like $\prod_{j}T_{j}$ where $T_{j}=U_{j}$ for finitely many $j$ and $T_{j}=X_{j}$ for all other $j$. this is precisely a set in $\beta$. Consequently, $\tau$ is a topology containing a basis for the topology $\tau_{\beta}$ hence it is finer than $\tau_{\beta}$.

Answer 2

Observe that if every projection is continuous with the topology $\tau$ then for every open set $U_j \subset X_j$ you have that $p_j^{-1}(U_j) \in \tau$.

Now, $$p_j^{-1}(U_j)=\left\{ f \in \prod_{i \in I}X_i: p_j(f) \in U_j \right\} = \left\{ f \in \prod_{i \in I}X_i: f(j) \in U_j \right\} = \prod_{i \in I}Y_i$$ where $Y_i=X_i$ if $i \not = j$ and $Y_j=U_j$.

As $\tau$ is a topology we have then that finite intersections of these sets will also be in $\tau$, but observe from the above that finite intersections of this sets are precisely the elements of $\beta$.

So, $\beta \subset \tau$ and, as $\beta$ is a basis for $\tau_\beta$ and $\tau$ is a topology, you get that $\tau_\beta \subset \tau$.

EDIT: I suggest you to investigate the concept of subbase and of initial topology, because it would be really useful to understand better the product topology.