Theorem 2.30
Let $\{(X_\alpha, \mathscr T_\alpha):\alpha \in \Lambda\}$ be an indexed family of topological spaces, and for each $\alpha \in \Lambda,$ let $\mathscr B_\alpha$ be a basis for $\mathscr T_\alpha$. Then the collection $\mathscr B$ of all sets of the form $\prod _{\alpha \in \Lambda}B_\alpha$, where $B_\alpha=X_\alpha$ for all but a finite number of members, $\beta_1,\beta_2,...,\beta_n$ of $\Lambda$ and $B_{\beta_i}\in \mathscr B_{\beta_i}$ for each $i=1,2,3,..,n,$ is the basis for the product topology on $\prod _{\alpha \in \Lambda}X_\alpha$.
My Doubt:- It is the theorem from the 'Foundation of Topology' by C.W Patty.
(1) By Theorem 2.30, it should be $U_\alpha \in \mathscr B_\alpha$ instead of $U_\alpha \in \mathscr T_\alpha$. Right?
(2) It should be $\pi_{\beta}(B)=X_\beta$. Right?
(3) Can you please explain the underlined statement?
How does the statement contradict the proposition?
No, $U_\alpha \in \mathcal{T}_\alpha$ is right. We know that for a local base element $B \in \mathcal{B}$ (there are no $\mathcal{B}_\alpha$ in this direction of the proof at all!) there is a standard-basic open subset that contains $x$ and sits inside $B$; this follows simply from sets of the described form being a base for the topology of the product (which is what 2.30 says when applied to $\mathcal{B}_\alpha = \mathcal{T}_\alpha$ for all $\alpha$ e.g.).
(2) says that for each $B$, almost all (all but finitely many) $\beta$ obey $\pi_\beta[B]=X_\beta$ (indeed, the $X_\alpha$ is a typo). So for all $B\in \mathcal{B}$ together we have at most countably many exceptions, so still plenty of such $\beta$ exist.
But then recall how $x$ was chosen: for all indices in $\Gamma$, so also for this $\beta$ with the property
$$\forall B \in \mathcal{B}: \pi_\beta[B]=X_\beta \tag{1}$$
we have a neighbourhood $x_\beta \in U_\beta \neq X_\beta$. So $x \in \pi_\beta^{-1}[U_\beta]$, which is an open neighbourhood of $x$ so as $\mathcal{B}$ is a local base at $x$, there must be a $B' \in \mathcal{B}$ such that
$$x \in B' \subseteq \pi_\beta^{-1}[U_\beta]$$ but his implies that $$X_\beta = \text{ by (1) } = \pi_\beta[B'] \subseteq U_\beta \neq X_\beta$$
which is a contradiction.