Characterization of Open Set by Open Cover
Jump to navigation
Jump to search
This article needs proofreading. Please check it for mathematical errors. If you believe there are none, please remove {{Proofread}} from the code.To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Proofread}} from the code. |
Theorem
Let $T = \struct{S, \tau}$ be a topological space.
Let $\UU$ be an open cover of $T$.
For each $U \in \UU$, let $\tau_U$ denote the subspace topology on $U$.
Let $W \subseteq S$.
Then $W$ is open in $T$ if and only if:
- $\forall U \in \UU: W \cap U$ is open in $\struct{U, \tau_U}$
Proof
Necessary Condition
This follows immediately from the definition of subspace topology.
$\Box$
Sufficient Condition
Let:
- $W \cap U$ be open in $\struct{U, \tau_U}$ for each $U \in \UU$
From Open Set in Open Subspace:
- $\forall U \in \UU : W \cap U$ is open in $T$
We have:
\(\ds W\) | \(=\) | \(\ds W \cap S\) | Intersection with Subset is Subset | |||||||||||
\(\ds \) | \(=\) | \(\ds W \cap \paren{\bigcup \UU}\) | Definition of Open Cover | |||||||||||
\(\ds \) | \(=\) | \(\ds \bigcup \set{W \cap U : U \in\UU}\) | Intersection Distributes over Union |
By Open Set Axiom $\paren {\text O 1 }$: Union of Open Sets:
- $W$ is open in $T$
$\blacksquare$