Box Topology may not be Coarsest Topology such that Projections are Continuous/Lemma

Theorem
Let $\struct{X, \tau}$ be a topological space.

Let $U \in \tau$ such that $U \neq \O$ and $U \neq X$.

Let:
 * $Y = \displaystyle \prod_{n \mathop \in \N } X = X \times X \times X \times \ldots$

be the countable Cartesian product of $\family {X}_{n \in \N}$.

Let $\tau_T$ be the Tychonoff topology on $Y$.

Let $\tau_b$ be the box topology on $Y$.

Let:
 * $V = \displaystyle \prod_{n \mathop \in \N } U = U \times U \times U \times \ldots$

be the countable Cartesian product of $\family {U}_{n \in \N}$.

Then:
 * $V$ is an element of the box topology $\tau_b$
 * $V$ is not an element of the Tychonoff topology $\tau_T$

Proof
By the definition of the box topology:
 * $V \in \tau_b$

Let $W \in \tau_T$.

Case 1
Let $W \cap V = \O$.

Since $U \neq \O$, there exists $x \in U$.

Then $\family {x}_{n \in \N} \in V$

So $\family {x}_{n \in \N} \notin W$

By the definition of set equality:
 * $V \neq W$

Case 2
Let $W \cap V \neq \O$.

From Natural Basis of Tychonoff Topology, $W = \displaystyle \prod_{n \mathop \in \N } W_n$ where:
 * for all $n \in \N : W_n \in \tau$
 * for all but finitely many indices $n : W_n = X$

Let $m = \max \set{n : W_n \neq X}$

Then:
 * $W_{m + 1} = X$

Since $W \cap V \neq \O$, there exists $\family {z_n} \in W \cap V$.

Since $U \neq X$, there exists $y \in X \setminus U$.

Let $\family {y_n}$ be defined as:
 * $y_n = \begin{cases}

z_n & : n \ne m + 1 \\ y & : n = m + 1 \end{cases}$

Since $y_{m + 1} = y \in X = W_{m + 1}$, it follows that:
 * $\family {y_n} \in W$

Since $y_{m + 1} = y \notin U$, it follows that:
 * $\family {y_n} \notin V$

By the definition of set equality:
 * $W \neq V$

In either case $W \neq V$.

Since $W$ was arbitrary, it follows that for all $W \in \tau_T$, $W \neq V$.

Thus $V \notin \tau_T$