Natural Basis of Product Topology/Lemma 2

From ProofWiki
Jump to navigation Jump to search

Lemma for Natural Basis of Product Topology

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

Let $X$ be the cartesian product of $\family {X_i}_{i \mathop \in I}$:

$\ds X := \prod_{i \mathop \in I} X_i$

Let $\BB$ be the set of cartesian products of the form $\ds\prod_{i \mathop \in I} U_i$ where:

for all $i \in I : U_i \in \tau_i$
for all but finitely many indices $i : U_i = X_i$


Then:

$\forall B_1, B_2 \in \BB : B_1 \cap B_2 \in \BB$


Proof

Let $B_1, B_2 \in \BB$.


Let $B_1 = \ds \prod_{i \mathop \in I} U_i$ where:

for all $i \in I : U_i \in \tau_i$
for all but finitely many indices $i : U_i = X_i$

Let $J_1$ be the finite set of indices such that:

$J_1 = \set {j \in I : U_i \neq X_i}$


Let $B_2 = \ds \prod_{i \mathop \in I} V_i$ where:

for all $i \in I : V_i \in \tau_i$
for all but finitely many indices $i : V_i = X_i$

Let $J_2$ be the finite set of indices such that:

$J_2 = \set{j \in I : V_i \neq X_i}$


Then:

\(\ds B_1 \cap B_2\) \(=\) \(\ds \paren{\prod_{i \mathop \in I} U_i } \cap \paren{\prod_{i \mathop \in I} V_i}\)
\(\ds \) \(=\) \(\ds \prod_{i \mathop \in I} \paren{ U_i \cap V_i}\) General Case of Cartesian Product of Intersections


Now:

\(\ds \forall i \in I: \, \) \(\ds U_i, V_i \in \tau_i\) \(\leadsto\) \(\ds U_i \cap V_i \in \tau_i\) Definition of topology $\tau_i$

and

\(\ds \forall i \in I \setminus \paren {J_1 \cup J_2}: \, \) \(\ds U_i \cap V_i\) \(=\) \(\ds X_i \cap X_i\) Definition of $J_1$ and $J_2$
\(\ds \) \(=\) \(\ds X_i\) Set Intersection is Idempotent

From Finite Union of Finite Sets is Finite, $J_1 \cup J_2$ is finite.

Then:

for all but finitely many indices $i : U_i \cap V_i = X_i$

Thus $B_1 \cap B_2 \in \BB$

$\blacksquare$