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[parent] every net has a universal subnet (Theorem)

Theorem - Let $ X$ be a non-empty set. Every net $ (x_{\alpha})_{\alpha \in \mathcal{A}}$ in $ X$ has a universal subnet. That is, there is a subnet such that for every $ E \subseteq X$ either the subnet is eventually in $ E$ or eventually in $ X-E$.

Proof : Let $ \mathcal{F}$ be a section filter for the net $ (x_{\alpha})_{\alpha \in \mathcal{A}}$.

Let $ \mathcal{D}=\{(\alpha,U):\alpha \in \mathcal{A}\;,\; U \in \mathcal{F},\; x_{\alpha} \in U \}$. $ \mathcal{D}$ is a directed set under the order relation given by

$\displaystyle (\alpha,U) \leq (\beta,V) \Longleftrightarrow \begin{cases} \alpha \leq \beta \ V \subseteq U \end{cases}$

The map $ f:\mathcal{D} \longrightarrow \mathcal{A}$ defined by $ f(\alpha,U):=\alpha$ is order preserving and cofinal. Therefore there is a subnet $ (y_{(\alpha,U)})_{(\alpha,U) \in \mathcal{D}}$ of $ (x_{\alpha})_{\alpha \in \mathcal{A}}$ associated with the map $ f$ (that is, $ y_{(\alpha,U)} = x_{\alpha}$).

We now prove that $ (y_{(\alpha,U)})_{(\alpha,U) \in \mathcal{D}}$ is a universal net.

Let $ E \subseteq X$. We have that $ (y_{(\alpha,U)})_{(\alpha,U) \in \mathcal{D}}$ is frequently in $ E$ or frequently in $ X-E$.

Suppose $ (y_{(\alpha,U)})_{(\alpha,U) \in \mathcal{D}}$ is frequently in $ E$.

Let $ A \in \mathcal{F}$ and $ S(\alpha):=\{x_{\beta}:\alpha \leq \beta\}$. We have that $ S(\alpha) \in \mathcal{F}$ by definition of section filter.

As $ \mathcal{F}$ is a filter, $ A \cap S(\alpha) \neq \emptyset$ and so there exists $ \beta$ with $ \alpha \leq \beta$ such that $ x_{\beta} \in A$. Hence, $ (\beta,A) \in \mathcal{D}$.

As $ (y_{(\alpha,U)})_{(\alpha,U) \in \mathcal{D}}$ is frequently in $ E$, there exists $ (\gamma,B) \in \mathcal{D}$ with $ (\beta,A) \leq (\gamma,B)$ such that $ y_{(\gamma,B)} \in E$.

Also, $ y_{(\gamma,B)}$ is in $ B$, and therefore, in $ A$. So $ A \cap E \neq \emptyset$.

We conclude that $ E \cap A \neq \emptyset$ for every $ A \in \mathcal{F}$. Therefore, $ \mathcal{F} \cup \{E\}$ generates a filter in $ X$. As $ \mathcal{F}$ is a maximal filter we conclude that $ E \in \mathcal{F}$, and consequently, $ (\gamma,E) \in \mathcal{D}$.

We can now see that for every $ (\delta,C)$ with $ (\gamma,E) \leq (\delta,C)$, $ y_{(\delta,C)}$ is in $ C$ and so is in $ E$. Therefore, $ (y_{(\alpha,U)})_{(\alpha,U) \in \mathcal{D}}$ is eventually in $ E$.

Remark: If $ (y_{(\alpha,U)})_{(\alpha,U) \in \mathcal{D}}$ is frequently in $ X-E$, by an analogous argument we can conclude that it is eventually in $ X-E$.

This proves that $ (y_{(\alpha,U)})_{(\alpha,U) \in \mathcal{D}}$ is a universal subnet of $ (x_{\alpha})_{\alpha \in \mathcal{A}}$. $ \square$



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See Also: ultranet


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Cross-references: maximal filter, filter, frequently in, cofinal, map, relation, order, directed set, section filter, proof, eventually, subnet, net
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This is version 4 of every net has a universal subnet, born on 2007-07-25, modified 2007-07-25.
Object id is 9795, canonical name is EveryNetHasAUniversalSubnet.
Accessed 778 times total.

Classification:
AMS MSC54A20 (General topology :: Generalities :: Convergence in general topology )

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