I am reading the paper Fuzzy Topological Spaces and Fuzzy Compactness by Robert Lowen. I have proved the theorem 2.2: $(X,\delta)$ is topologically generated if and only if for each continuous function $f \in \mathscr{G}(I,I_r)$ and for each $\nu \in \delta$, $f \circ \nu \in \delta$.

Assume $(X,\delta)$ is topologically generated. Since $\nu \in \mathscr{G}(\mathcal{J},I_e) = \delta$ and $f \in \mathscr{G}(I_r,I_r)$, then $f \circ \nu \in \mathscr{G}(\mathcal{J},I_r) = \delta$.

Conversely, assume $\mu \in \bar{\delta}$. Recall that $\bar{\delta} = \omega \circ \iota (\delta)$. This shows $\mu \in \mathscr{G}\big(\iota(\delta),I_r\big)$. Since a basis for $\iota(\delta)$ is provided by the finite intersections \begin{align*} \bigcap_{i=1}^n \nu_i^{-1} \big((r_i,1]\big)\;\;\; \text{ for some }\nu_i \in \delta,\,r_i \in I; \end{align*} this is equivalent to saying for any $r \in I$, any $x \in \mu^{-1}\big((r,1]\big)$, $(r,1]$ is open in $I_r$ and $\mu^{-1}(r,1]$ is open in $\iota(\delta)$ since $\mu \in \mathscr{G}\big(\iota(\delta),I_r\big)$. Hence, for every $x \in \mu^{-1}(r,1]$, there exists finite open set $(r_i,1]$ such that \begin{align*} x \in \bigcap_{i \in I_{r,x}} \nu_i^{-1} \big((r_i,1]\big) \subseteq \mu_i^{-1} \big((r_i,1]\big). \end{align*} Now, we want to show $\mu$ is closed under some finite intersection and arbitrary union of basis of $\delta$. Fix $x$ and let $\mu(x) = k_x \in (r,1]$, then $\forall x < k_x,\, \exists$ a finite index set $I_{r}$ such that \begin{align*} x \in \bigcap_{i \in I_{r}} \nu_i^{-1} \big((r_i,1]\big) \subseteq \mu_i^{-1} \big((r_i,1]\big). \end{align*} Then, $\forall r < k_x$ and $\forall i \in I_r$, let \begin{align*} \mu_{i,r}(y) =\big( (r \chi_{\scriptscriptstyle{(i_i,1]}}) \circ \nu_i\big) (y) = \begin{cases} r,\, \text{if } \nu_i(y) > r_i,\\ 0,\,\text{if } \nu_i(y) \leq r_i. \end{cases} \end{align*} where $\mu_{i,r} \in \delta$ and $r \chi_{\scriptscriptstyle{(i_i,1]}} \in \mathscr{G}(I_r,I_r)$. This indeed follows from our assumption $f \circ \nu $ for all $ f \in \mathscr{G}(I_r,I_r) \text{ and } \nu \in \delta$ Then, let $\nu_{r}^x = \displaystyle\inf_{i \in I_{r}} \mu_{i,r} \in \delta$. Since $I_r$ is finite and hence $\nu_r^x = \displaystyle\min_{i \in I_r} \mu_{i,r}$, then we have \begin{align*} \nu_{r}^x(y) = \begin{cases} r,\,\text{if } \;\forall i \in I_{r}\text{ we have }\nu_i(y)>r_i,\\ 0,\,\text{if }\;\exists j \in I_{r}\text{ such that }\nu_i(y) \leq r_j. \end{cases} \end{align*} Hence if $\nu_{r}^x(y) = r $, then $\nu_i(y) > r_i$ for all $i \in I_r$. Since \begin{align*} y \in \bigcap_{i=1}^n \nu_i^{-1}(r_i,1] \subseteq \mu^{-1}(r,1], \end{align*}therefore for every $y \in [0,1]$, $\mu(y) > r$ which shows $\mu \geq \nu_r^x$ for every $x \in \mu^{-1}(r,1]$ and $r < k_x$. Now, it is easily to seen that \begin{align*} \mu = \sup_{x \in X} \sup_{r < k_x} \nu_{r}^x(y) \in \delta. \end{align*} Hence, if every $\mu \in \bar{\delta}$, then $\mu \in \delta$. This shows $\bar{\delta} = \delta$ which implies $(X,\delta)$ is topologically generated.

I began by assuming that $\mu \in \bar{\delta}$ where $\bar{\delta}$ is a topologically generated fuzzy topology. Then I constructed a basis from $\delta$ and showed that $\mu$ can be expressed as a finite intersection and arbitrary union of the open sets in this basis. My final goal is to show if $\mu \in \bar{\delta}$, then $\mu \in \delta$, which would shows $(X,\delta)$ is topologically generated. Could someone help me verify if my proof is correcct?

I am reading the paper Fuzzy Topological Spaces and Fuzzy Compactness by Robert Lowen. I have proved the theorem 2.2: $(X,\delta)$ is topologically generated if and only if for each continuous function $f \in \mathscr{G}(I,I_r)$ and for each $\nu \in \delta$, $f \circ \nu \in \delta$.

Assume $(X,\delta)$ is topologically generated. Since $\nu \in \mathscr{G}(\mathcal{J},I_e) = \delta$ and $f \in \mathscr{G}(I_r,I_r)$, then $f \circ \nu \in \mathscr{G}(\mathcal{J},I_r) = \delta$.

Conversely, assume $\mu \in \bar{\delta}$. Recall that $\bar{\delta} = \omega \circ \iota (\delta)$. This shows $\mu \in \mathscr{G}\big(\iota(\delta),I_r\big)$. Since a basis for $\iota(\delta)$ is provided by the finite intersections \begin{align*} \bigcap_{i=1}^n \nu_i^{-1} \big((r_i,1]\big)\;\;\; \text{ for some }\nu_i \in \delta,\,r_i \in I; \end{align*} this is equivalent to saying for any $r \in I$, any $x \in \mu^{-1}\big((r,1]\big)$, $(r,1]$ is open in $I_r$ and $\mu^{-1}(r,1]$ is open in $\iota(\delta)$ since $\mu \in \mathscr{G}\big(\iota(\delta),I_r\big)$. Hence, for every $x \in \mu^{-1}(r,1]$, there exists finite open set $(r_i,1]$ such that \begin{align*} x \in \bigcap_{i \in I_{r,x}} \nu_i^{-1} \big((r_i,1]\big) \subseteq \mu_i^{-1} \big((r_i,1]\big). \end{align*} Now, we want to show $\mu$ is closed under some finite intersection and arbitrary union of basis of $\delta$. Fix $x$ and let $\mu(x) = k_x \in (r,1]$, then $\forall x < k_x,\, \exists$ a finite index set $I_{r}$ such that \begin{align*} x \in \bigcap_{i \in I_{r}} \nu_i^{-1} \big((r_i,1]\big) \subseteq \mu_i^{-1} \big((r_i,1]\big). \end{align*} Then, $\forall r < k_x$ and $\forall i \in I_r$, let \begin{align*} \mu_{i,r}(y) =\big( (r \chi_{\scriptscriptstyle{(i_i,1]}}) \circ \nu_i\big) (y) = \begin{cases} r,\, \text{if } \nu_i(y) > r_i,\\ 0,\,\text{if } \nu_i(y) \leq r_i. \end{cases} \end{align*} where $\mu_{i,r} \in \delta$ and $r \chi_{\scriptscriptstyle{(i_i,1]}} \in \mathscr{G}(I_r,I_r)$. This indeed follows from our assumption $f \circ \nu $ for all $ f \in \mathscr{G}(I_r,I_r) \text{ and } \nu \in \delta$. Then, let $\nu_{r}^x = \displaystyle\inf_{i \in I_{r}} \mu_{i,r} \in \delta$. Since $I_r$ is finite and hence $\nu_r^x = \displaystyle\min_{i \in I_r} \mu_{i,r}$, then we have \begin{align*} \nu_{r}^x(y) = \begin{cases} r,\,\text{if } \;\forall i \in I_{r}\text{ we have }\nu_i(y)>r_i,\\ 0,\,\text{if }\;\exists j \in I_{r}\text{ such that }\nu_i(y) \leq r_j. \end{cases} \end{align*} Hence if $\nu_{r}^x(y) = r $, then $\nu_i(y) > r_i$ for all $i \in I_r$. Since \begin{align*} y \in \bigcap_{i=1}^n \nu_i^{-1}(r_i,1] \subseteq \mu^{-1}(r,1], \end{align*}therefore for every $y \in [0,1]$, $\mu(y) > r$ which shows $\mu \geq \nu_r^x$ for every $x \in \mu^{-1}(r,1]$ and $r < k_x$. Now, it is easily to seen that \begin{align*} \mu = \sup_{x \in X} \sup_{r < k_x} \nu_{r}^x(y) \in \delta. \end{align*} Hence, if every $\mu \in \bar{\delta}$, then $\mu \in \delta$. This shows $\bar{\delta} = \delta$ which implies $(X,\delta)$ is topologically generated.

I began by assuming that $\mu \in \bar{\delta}$ where $\bar{\delta}$ is a topologically generated fuzzy topology. Then I constructed a basis from $\delta$ and showed that $\mu$ can be expressed as a finite intersection and arbitrary union of the open sets in this basis. My final goal is to show if $\mu \in \bar{\delta}$, then $\mu \in \delta$, which would shows $(X,\delta)$ is topologically generated. Could someone help me verify if my proof is correcct?

I am reading the paper Fuzzy Topological Spaces and Fuzzy Compactness by Robert Lowen. I got stuck at theorem 2.2:  $(X,\delta)$ is topologically generated if and only if for each continuous function $f \in \mathscr{G}(I,I_r)$ and for each $\nu \in \delta$, $f \circ \nu \in \delta$.

1. For the only if part, the author says it is trivial but I have no idea on it. Can someone give me some hints?

2. For the if only part,

Suppose $\mu \in \bar{\delta}$. Since a basis for $\iota(\delta)$ is provided by the finite intersections \begin{align*} \bigcap_{i=1}^n \nu_i^{-1} \big((\epsilon_i,1]\big)\;\;\; \text{ where }\nu_i \in \delta,\,\epsilon_i \in I; \end{align*} this is equivalent to saying for all $\epsilon \in I$ and for all $x \in \mu^{-1}\big((\epsilon,1]\big)$, there exists a finite set $I_{\epsilon,x}$ such that \begin{align*} x \in \bigcap_{i \in I_{\epsilon,x}} \nu_i^{-1} \big((\epsilon_i,1]\big) \subseteq \mu_i^{-1} \big((\epsilon_i,1]\big). \end{align*} Now, fix $x$ and let $\mu(x) = k_x$, then $\forall x < k_x,\, \exists$ a finite set $I_{\epsilon}$ such that \begin{align*} x \in \bigcap_{i \in I_{\epsilon}} \nu_i^{-1} \big((\epsilon_i,1]\big) \subseteq \mu_i^{-1} \big((\epsilon_i,1]\big). \end{align*} Then, $\forall \epsilon < k_x$ and $\forall i \in I_{\epsilon}$, let \begin{align*} \mu_{i,\epsilon}(y) =\big( (\epsilon \chi_{\scriptscriptstyle{(\epsilon_i,1]}}) \circ \nu_i\big) (y) = \begin{cases} \epsilon,\, \text{if } \nu_i(y) > \epsilon_i,\\ 0,\,\text{if } \nu_i(y) \leq \epsilon_i. \end{cases} \end{align*} where $\mu_{i,\epsilon} \in \delta$. Then, let $\nu_{\epsilon}^x = \displaystyle\inf_{i \in I_{\epsilon}} \mu_{i,\epsilon} \in \delta$ then we have \begin{align*} \nu_{\epsilon}^x(y) = \begin{cases} \epsilon,\,\text{if } \;\forall i \in I_{\epsilon}\text{ we have }\nu_i(y)>\epsilon_i,\\ 0,\,\text{if }\;\exists j \in I_{\epsilon}\text{ such that }\nu_i(y) \leq \epsilon_j. \end{cases} \end{align*} Hence $\nu_{\epsilon}^x(y) = \epsilon \Rightarrow \mu(y) > \epsilon$ and $\forall \epsilon < k_x, \nu_{\epsilon}^x(y)\leq \mu$. Now, it is easily to seen that \begin{align*} \mu = \sup_{x \in X} \sup_{\epsilon < k_x} \nu_{\epsilon}^x(y) \in \delta. \end{align*}

What I understand for this part is it construct a basis for $\delta$ from finite intersections of inverse image of $(\epsilon_i,1]$ under functions $\nu_i \in \delta$. This ensure each point $x \in (X,\delta)$ has a neighborhood in $\delta$. Then by $\mu_{i,r}(y)$ and $\nu_r^x(y)$ based on the values of $\nu_i$ over finite sets, it make sure these neighborhoods cover all points in $\delta$ and are closed under composition with continuous function. Then the supremum make sure $\delta$ includes all needed open sets to form a topology, which shows $(X,\delta)$ is topologically generated. Can anyone help me to check my understand correct or not?

I am reading the paper Fuzzy Topological Spaces and Fuzzy Compactness by Robert Lowen. I have proved the theorem 2.2:   $(X,\delta)$ is topologically generated if and only if for each continuous function $f \in \mathscr{G}(I,I_r)$ and for each $\nu \in \delta$, $f \circ \nu \in \delta$.

Assume $(X,\delta)$ is topologically generated. Since $\nu \in \mathscr{G}(\mathcal{J},I_e) = \delta$ and $f \in \mathscr{G}(I_r,I_r)$, then $f \circ \nu \in \mathscr{G}(\mathcal{J},I_r) = \delta$.

Conversely, assume $\mu \in \bar{\delta}$. Recall that $\bar{\delta} = \omega \circ \iota (\delta)$. This shows $\mu \in \mathscr{G}\big(\iota(\delta),I_r\big)$. Since a basis for $\iota(\delta)$ is provided by the finite intersections \begin{align*} \bigcap_{i=1}^n \nu_i^{-1} \big((r_i,1]\big)\;\;\; \text{ for some }\nu_i \in \delta,\,r_i \in I; \end{align*} this is equivalent to saying for any $r \in I$, any $x \in \mu^{-1}\big((r,1]\big)$, $(r,1]$ is open in $I_r$ and $\mu^{-1}(r,1]$ is open in $\iota(\delta)$ since $\mu \in \mathscr{G}\big(\iota(\delta),I_r\big)$. Hence, for every $x \in \mu^{-1}(r,1]$, there exists finite open set $(r_i,1]$ such that \begin{align*} x \in \bigcap_{i \in I_{r,x}} \nu_i^{-1} \big((r_i,1]\big) \subseteq \mu_i^{-1} \big((r_i,1]\big). \end{align*} Now, we want to show $\mu$ is closed under some finite intersection and arbitrary union of basis of $\delta$. Fix $x$ and let $\mu(x) = k_x \in (r,1]$, then $\forall x < k_x,\, \exists$ a finite index set $I_{r}$ such that \begin{align*} x \in \bigcap_{i \in I_{r}} \nu_i^{-1} \big((r_i,1]\big) \subseteq \mu_i^{-1} \big((r_i,1]\big). \end{align*} Then, $\forall r < k_x$ and $\forall i \in I_r$, let \begin{align*} \mu_{i,r}(y) =\big( (r \chi_{\scriptscriptstyle{(i_i,1]}}) \circ \nu_i\big) (y) = \begin{cases} r,\, \text{if } \nu_i(y) > r_i,\\ 0,\,\text{if } \nu_i(y) \leq r_i. \end{cases} \end{align*} where $\mu_{i,r} \in \delta$ and $r \chi_{\scriptscriptstyle{(i_i,1]}} \in \mathscr{G}(I_r,I_r)$. This indeed follows from our assumption $f \circ \nu $ for all $ f \in \mathscr{G}(I_r,I_r) \text{ and } \nu \in \delta$ Then, let $\nu_{r}^x = \displaystyle\inf_{i \in I_{r}} \mu_{i,r} \in \delta$. Since $I_r$ is finite and hence $\nu_r^x = \displaystyle\min_{i \in I_r} \mu_{i,r}$, then we have \begin{align*} \nu_{r}^x(y) = \begin{cases} r,\,\text{if } \;\forall i \in I_{r}\text{ we have }\nu_i(y)>r_i,\\ 0,\,\text{if }\;\exists j \in I_{r}\text{ such that }\nu_i(y) \leq r_j. \end{cases} \end{align*} Hence if $\nu_{r}^x(y) = r $, then $\nu_i(y) > r_i$ for all $i \in I_r$. Since \begin{align*} y \in \bigcap_{i=1}^n \nu_i^{-1}(r_i,1] \subseteq \mu^{-1}(r,1], \end{align*}therefore for every $y \in [0,1]$, $\mu(y) > r$ which shows $\mu \geq \nu_r^x$ for every $x \in \mu^{-1}(r,1]$ and $r < k_x$. Now, it is easily to seen that \begin{align*} \mu = \sup_{x \in X} \sup_{r < k_x} \nu_{r}^x(y) \in \delta. \end{align*} Hence, if every $\mu \in \bar{\delta}$, then $\mu \in \delta$. This shows $\bar{\delta} = \delta$ which implies $(X,\delta)$ is topologically generated.

I began by assuming that $\mu \in \bar{\delta}$ where $\bar{\delta}$ is a topologically generated fuzzy topology. Then I constructed a basis from $\delta$ and showed that $\mu$ can be expressed as a finite intersection and arbitrary union of the open sets in this basis. My final goal is to show if $\mu \in \bar{\delta}$, then $\mu \in \delta$, which would shows $(X,\delta)$ is topologically generated. Could someone help me verify if my proof is correcct?