Subspace of Product Space Homeomorphic to Factor Space/Proof 1

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Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be a family of topological spaces where $I$ is an arbitrary index set.


$\displaystyle T = \struct {X, \tau} = \prod_{i \mathop \in I} \struct {X_i, \tau_i}$

Suppose that $X$ is non-empty.

Then for each $i \in I$ there is a subspace $Y_i \subseteq X$ which is homeomorphic to $T_i = \struct {X_i, \tau_i}$.

Specifically, for any $z \in X$, let:

$Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$

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

Then $\struct {Y_i, \upsilon_i}$ is homeomorphic to $\struct {X_i, \tau_i}$, where the homeomorphism is the restriction of the projection $\pr_i$ to $Y_i$.


Consider the restriction of the projection:

$\pr_i {\restriction_{Y_i} }: Y_i \to X_i$

From Projection from Product Topology is Continuous, $\pr_i {\restriction_{Y_i} }$ is continuous.

From Projection from Product Topology is Open, $\pr_i {\restriction_{Y_i} }$ is open.

$\pr_i {\restriction_{Y_i} }$ is also bijective.


Thus, by definition, we have that $\pr_i {\restriction_{Y_i} }: Y_i \to X_i$ is a homeomorphism.