Subspace of Product Space is Homeomorphic to Factor Space/Proof 1

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
By we have that $\operatorname{pr}_i {\restriction_{Y_i}}: Y_i \to X_i$ is a homeomorphism because it is open, continuous and bijective.
 * Projection from Product Topology is Continuous
 * Projection from Product Topology is Open