Obtaining Generalized Topology by constructing Generalized neighbourhood

Question

$X$ is a given set . To define Generalized Neighborhoods on $X$ Consider $ P ( P (X))$ - the power set of the power set of $X.$

Let $$\psi : X \rightarrow P(P(X))$$ satisfying $x\in V$ for $V\in \psi(x)$. Then $V\in \psi(x)$ is called a Generalized Neighborhood of x.

Now here is my problem with this :

if $$X=\{a,b\}$$ Then $$P(X)=\{\Phi,\{a\},\{b\},X\}$$ And $$P(P(X))=\{\{\Phi\},\{\{a\}\},\{\{b\}\},\{\Phi,\{a\} \},\{\Phi,\{b\} \},\{\{b\},\{a\}\},\{\Phi,\{X\} \},\{\{X\},\{a\} \},\{\{X\},\{b\} \} \}$$ $-$ Is this correct $?$

Now $$\psi : X \rightarrow P(P(X))$$ being the function and looking at the range set of it , I see it is not possible for $a$ or $b$ to belong to $\psi(a)$ or $\psi(b)$.

What can actually belong to $\psi(a)$ or $\psi(b)$ are $\{a\}$ or $\{b\}$. So , what is this $?$ I cannot understand the definition and cannot move further . Please help.

Answer 1

The elements of $\wp(\wp(X))$ are the subsets of $\wp(X)=\big\{\varnothing,\{a\},\{b\},X\big\}$. Since $\wp(X)$ has $4$ elements, it has $2^4=16$ subsets. They are:

$$\begin{align*} &\varnothing,\\ &\{\varnothing\},\big\{\{a\}\big\},\big\{\{b\}\big\},\{X\},\\ &\big\{\varnothing,\{a\}\big\},\big\{\varnothing,\{b\}\big\},\{\varnothing,X\},\big\{\{a\},\{b\}\big\},\big\{\{a\},X\big\},\big\{\{b\},X\big\},\\ &\big\{\varnothing,\{a\},\{b\}\big\},\big\{\varnothing,\{a\},X\big\},\big\{\varnothing,\{b\},X\big\},\big\{\{a\},\{b\},X\big\},\\ &\big\{\varnothing,\{a\},\{b\},X\big\}, \end{align*}$$

where for clarity I’ve divided them up according to their cardinalities. Let’s see which of these $16$ families of subsets of $X$ can be $\psi(a)$.

The definition says that if $\psi(a)$ must be one of these $16$ families of subsets of $X$, and moreover $a$ must be an element of every member of $\psi(a)$ (not an element of $\psi(a)$ itself). This immediately tells us that $\varnothing$ cannot be one of the members of $\psi(a)$, because $a\notin\varnothing$. It also tells us that $\{b\}$ cannot be one of the members of $\psi(a)$, because $a\notin\{b\}$. Thus, the only possible members of $\psi(a)$ are the sets $\{a\}$ and $X$: these are the only subsets of $X$ that contain $a$. Thus, of the $16$ subsets of $\wp(X)$ that I listed above, the only ones that can possible be $\psi(a)$ are

$$\begin{align*} &\varnothing\;,\\ &\big\{\{a\}\big\}\;,\\ &\{X\}\;,\text{ and}\\ &\big\{\{a\},X\big\}\;. \end{align*}\tag{1}$$

Each of these four families of subsets of $X$ has the property that if $V$ is a member of the family, then either $V=\{a\}$ or $V=X$, and in either case $a\in V$.

Note: In the case of $\varnothing$ this statement is vacuously true, simply because there is no $V\in\varnothing$ at all, let alone one that does not contain $a$. I’d have expected the definition to require that $\psi(x)$ be a non-empty subset of $\wp(X)$ for each $x\in X$. If that restriction is added, the only possibilities for $\psi(a)$ in your example are the last three families listed in $(1)$.

If $\psi(a)=\big\{\{a\}\big\}$, we’re saying that $\{a\}$ is the only generalized nbhd of $a$ in the system $\psi$. If $\psi(a)=\{X\}$, $X=\{a,b\}$ is the only generalized nbhd of $a$ in the system $\psi$; and if $\psi(a)=\big\{\{a\},X\big\}$, we’re saying that $\{a\}$ and $X$ are both generalized nbhds of $a$ in the system $\psi$. Since a generalized nbhd of $a$ is required to contain $a$ as an element, these are the only possible systems of generalized nbhds at $a$, provided that we require $a$ to have at least one generalized nbhd (so that $\psi(a)\ne\varnothing$).