Let $X$ a finite set, and $X^{*}=X\cup \{\omega\}$ wiht $\omega\notin X$. Given a filter $\mathcal{F}$ on $X$, Show that

$$\mathcal{T}(\mathcal{F}):=2^{X}\cup\{F\cup \omega\mid F\in\mathcal{F}\}$$ is a topology for $X^{*}$

That i have tried:

$(i)$ We have $\emptyset$ and $X\in 2^{X}$, then $\emptyset$ and $X\in \mathcal{T}(\mathcal{F})$

$(ii)$Let $\{A_i\}_{i\in I}$ a family of element of $\mathcal{T}(\mathcal{F})$ then

$A_i\in\mathcal{T}(\mathcal{F})$ for all $i\in I$, so $A_i=K_i\cup F_i\cup\{\omega\}$ where $K_i\in2^X$ and $F_i\in\mathcal{F}$ for all $i\in I$.

as we know $\bigcup K_i\in \wp(X)$ and $\bigcup F_i\in\mathcal{T}$,then we have

$\bigcup A_i=\bigcup_{i\in I}\{K_i\cup F_i\cup\{\omega\}\}$ =$\{\bigcup_{i\in I}\{K_i\cup \bigcup_{i\in I} F_i\cup\{\omega\}\}$ then $\bigcup A_i\in \mathcal{T}(\mathcal{F})$.

$(iii)$ Let $I_n=\{1,2,3,\cdots,n\}$ a finite set and $\{A_j\}_{j\in I_n}$ a finite family of element $\mathcal{T}$ thus $A_j\in \mathcal{T}(\mathcal{F})$ where $A_j=K_j\cup F_j\cup\{\omega\}$, since we have $K_j\in\wp(X)$ and $F_j\in \mathcal{F}$, then $\bigcap_{j\in I_n}K_j\in\wp(x)$ and $\bigcap_{j\in I_n}F_j\in\mathcal{F}$ because $\mathcal{F}$ have the FIP (Finite intersection propiety).

Then we can say that

$$\bigcap_{j=1}^{n}A_j=\bigcap_{j=1}^{n}(K_j\cup F_j\cup\{\omega\})= \left(\bigcap^{n}_{j\in I_n}K_j\right)\cup\left(\bigcap^{n}_{j\in I_n}F_j\right)\cup\{\omega\}$$

Then we have that

$$A_j=\bigcap^{n}_{j=1}A_i\in \mathcal{T}(\mathcal{F})$$.

For $(i),(ii)$ and $(iii)$, $\mathcal{T}(\mathcal{F})$ is a topology.

Am I right?

  • 1
    $\begingroup$ I don't think the finite intersection of a union formula you use is correct. Also you seem to confuse $\omega$ and $\{\omega\}$. $\endgroup$ – Henno Brandsma May 11 '14 at 19:07
  • $\begingroup$ Also you need $X^\ast \in \mathcal{T}(\mathcal{F})$ $\endgroup$ – Henno Brandsma May 11 '14 at 19:09
  • $\begingroup$ Are you sure that $X$ is a finite set? $\endgroup$ – user642796 May 12 '14 at 7:05

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