Correspondence between Neighborhood Space and Topological Space

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Theorem

Let $S$ be a set.

Let $\left({S, \tau}\right)$ be a topological space.

Let $\left({S, \mathcal N}\right)$ be the neighborhood space induced by $\tau$ on $S$.

Let $\phi: \left({S, \tau}\right) \to \left({S, \mathcal N}\right)$ be the mapping defined as:

$\forall x \in S: \phi \left({x}\right) = x$
$\forall U \in \tau: \phi \left({U}\right) = U \in \mathcal N$


Let $\left({T, \mathcal N'}\right)$ be a neighborhood space.

Let $\left({T, \tau'}\right)$ be the topological space induced by $\mathcal N$ on $S$.

Let $\psi: \left({T, \mathcal N'}\right) \to \left({T, \tau'}\right)$ be the mapping defined as:

$\forall y \in T: \psi \left({y}\right) = y$
$\forall V \in \mathcal N': \psi \left({V}\right) = V \in \tau'$


Then:

$(1): \quad \phi^{-1} \left({\phi \left({S, \tau}\right)}\right) = \left({S, \tau}\right)$
$(2): \quad \psi^{-1} \left({\psi \left({T, \mathcal N'}\right)}\right) = \left({T, \mathcal N'}\right)$


Proof

From the construction of:

the neighborhood space induced by $\tau$ on $S$
the topological space induced by $\mathcal N$ on $S$

the mappings $\phi$ and $\psi$ are well-defined mappings.


From Topological Space induced by Neighborhood Space induced by Topological Space, $\phi$ is a bijection.

From Neighborhood Space induced by Topological Space induced by Neighborhood Space, $\psi$ is a bijection.

Hence the result.

$\blacksquare$


Sources