Subspace of Product Space is Homeomorphic to Factor Space/Proof 2

Theorem
Let $\left \langle {\left({X_i, \tau_i}\right)}\right \rangle_{i \mathop \in I}$ be a family of topological spaces where $I$ is an arbitrary index set.

Consider:
 * $\displaystyle T = \left({X, \tau}\right) = \prod_{i \mathop \in I} \left({X_i, \tau_i}\right)$

Suppose that $X$ is non-empty.

Then for each $i \in I$ there is a subspace $Y_i\subseteq X$ which is homeomorphic to $T$.

Specifically, for any $z \in X$, let:
 * $Y_i = \left\{{x \in X: \forall j \in I \setminus \left\{{i}\right\}: x_j = z_j}\right\}$

and let $\upsilon_i$ be the subspace topology of $Y_i$ relative to $\tau$.

Then $\left({Y_i,\upsilon_i}\right)$ is homeomorphic to $\left({X_i, \tau_i}\right)$, where the homeomorphism is the restriction of $\operatorname{pr}_i$ to $Y_i$.

Proof
For each $i \in I$, let $p_i = \operatorname{pr}_i {\restriction_{Y_i}}$.

By Projection from Product Topology is Continuous and Restriction of Continuous Mapping is Continuous: $p_i$ is continuous.

Let $U \in \upsilon$.

Then by the definition of the subspace topology, there is a $U' \in \tau$ such that $U = U' \cap Y_i$.

Thus for each $y \in Y_i$, there is a finite subset $I_y$ of $I$ and for each $k\in I_y$ there is a $V_k \in \tau_k$ such that $\displaystyle y \in \bigcap \operatorname{pr}_i^{-1} (V_k) \subseteq U'$.