Topological Product with Singleton

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Theorem

Let $T_1$ and $T_2$ be topological spaces.

Let $a \in T_1, b \in T_2$.

Let $T_1 \times T_2$ be the product space of $T_1$ and $T_2$.

Then:

$T_1$ is homeomorphic to the subspace $T_1 \times \left\{{b}\right\}$ of $T_1 \times T_2$
$T_2$ is homeomorphic to the subspace $\left\{{a}\right\} \times T_2$ of $T_1 \times T_2$


Proof

The conclusions are symmetrical.

Without loss of generality, therefore, it will be shown that $T_1$ is homeomorphic to the subspace $T_1 \times \left\{{b}\right\}$ of $T_1 \times T_2$.

Let $f: T_1 \to T_1 \times \set{b}$ be defined as:

$\map f x = \tuple {x, b}$

$f$ is clearly a bijection.


$f^{-1}$ is a restriction to the subspace of the projection of $T_1 \times T_2$ onto $T_1$.

This is continuous.

Therefore $f^{-1}$ is continuous.


It is to be shown that $f$ is continuous.

Let $x \in T_1$.

Let $U$ be an open set in $T_1 \times \set {b}$ such that:

$\map f x \in U$

Then for some $U'$ open in $T_1 \times T_2$:

$U' \cap \paren {T_1 \times \set {b}}= U$

By the definition of the product topology, there exist open sets $V_1$ and $V_2$ in $T_1$ and $T_2$, respectively, such that:

$\map f x = \tuple {x, b} \in V_1 \times V_2 \subseteq U'$

Then for any $y \in V_1$:

$\map f y = \tuple {y, b} \in V_1 \times V_2 \subseteq U'$

But:

$\tuple {y,b} \in T_1 \times \set{b}$

so:

$\tuple{y, b} \in U$

Thus if $y \in V_1$, it follows that $\map f y \in U$.


A Generalization


Let $X_a$ be a topological space for each $a \in I$.

Let:

$\displaystyle X = \prod_{a \mathop \in I} X_a$

Let $p_a$ be the $a$th projection for each $a \in I$.

Let $x_0 \in X$.

Let $k \in I$.

Let:

$S = \set {x \in X: \map {p_a} x = \map {p_a} {x_0} \text{ when } a \ne k}$

Then $S$ is homeomorphic to $X_k$.


Proof