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 topological product 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 $f: T_1 \to T_1 \times \left\{{b}\right\}$ be defined as:

$f \left({x}\right) = \left({x, b}\right)$

$f$ 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 $T_1 \times \left\{{b}\right\}$ such that:

$f \left({x}\right) \in U$

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

$U' \cap \left({T_1 \times \left\{{b}\right\}}\right) = U$

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:

$f \left({x}\right) = \left({x, b}\right) \in V_1 \times V_2 \subseteq U'$

Then for any $y \in V_1$:

$f \left({y}\right) = \left({y, b}\right) \in V_1 \times V_2 \subseteq U'$

But:

$\left({y,b}\right) \in T_1 \times \left\{{b}\right\}$

so:

$\left({y, b}\right) \in U$

Thus if $y \in V_1$, it follows that $f \left({y}\right) \in U$.


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:

$S = \left\{ {x \in X: p_a \left({x}\right) = p_a \left({x_0}\right) \text{ when } a \ne k}\right\}$

Then $S$ is homeomorphic to $X_k$.


Proof