Post History
#3: Post edited
by
Snoopy
·
2023-02-04T02:55:14Z (almost 2 years ago)
- When topologies are given by bases, one has a useful criterion in terms of the bases for determining whether one topology is finer than another:
- > Lemma 13.3. (Munkres's Topology p.81) Let $\mathscr{B}$ and $\mathscr{B}^{\prime}$ be bases for the topologies $\mathcal{T}$ and $\mathcal{T}^{\prime}$, respectively, on $X$. Then the following are equivalent:
- >
- > (1) $\mathcal{T}^{\prime}$ is finer than $\mathcal{T}$.
- > (2) For each $x \in X$ and each basis element $B \in \mathscr{B}$ containing $x$, there is a basis element $B^{\prime} \in \mathscr{B}^{\prime}$ such that $x \in B^{\prime} \subset B$.
- If one modifies condition (2) as:
- > (2') For each nonempty basis element $B \in \mathscr{B}$ there is a nonempty basis element $B^{\prime} \in \mathscr{B}^{\prime}$ such that $B^{\prime} \subset B$.
- then it is unlikely that one can conclude (2'): one cannot use the original argument for showing the implication (2) $\Rightarrow$ (1) anymore.
- Can anyone give a counter-example for (2') $\Rightarrow$ (1)?
- When topologies are given by bases, one has a useful criterion in terms of the bases for determining whether one topology is finer than another:
- > Lemma 13.3. (Munkres's Topology p.81) Let $\mathscr{B}$ and $\mathscr{B}^{\prime}$ be bases for the topologies $\mathcal{T}$ and $\mathcal{T}^{\prime}$, respectively, on $X$. Then the following are equivalent:
- >
- > (1) $\mathcal{T}^{\prime}$ is finer than $\mathcal{T}$.
- > (2) For each $x \in X$ and each basis element $B \in \mathscr{B}$ containing $x$, there is a basis element $B^{\prime} \in \mathscr{B}^{\prime}$ such that $x \in B^{\prime} \subset B$.
- If one modifies condition (2) as:
- > (2') For each nonempty basis element $B \in \mathscr{B}$ there is a nonempty basis element $B^{\prime} \in \mathscr{B}^{\prime}$ such that $B^{\prime} \subset B$.
- then it is unlikely that one can conclude (2'): one cannot use the original argument for showing the implication (2) $\Rightarrow$ (1) anymore.
- Can anyone give a counter-example for (2') $\Rightarrow$ (1)?
- [(2') edited.]
#2: Post edited
by
Snoopy
·
2023-02-03T20:25:48Z (almost 2 years ago)
The condition (2') edited.
- When topologies are given by bases, one has a useful criterion in terms of the bases for determining whether one topology is finer than another:
- > Lemma 13.3. (Munkres's Topology p.81) Let $\mathscr{B}$ and $\mathscr{B}^{\prime}$ be bases for the topologies $\mathcal{T}$ and $\mathcal{T}^{\prime}$, respectively, on $X$. Then the following are equivalent:
- >
- > (1) $\mathcal{T}^{\prime}$ is finer than $\mathcal{T}$.
- > (2) For each $x \in X$ and each basis element $B \in \mathscr{B}$ containing $x$, there is a basis element $B^{\prime} \in \mathscr{B}^{\prime}$ such that $x \in B^{\prime} \subset B$.
- If one modifies condition (2) as:
> (2') For each basis element $B \in \mathscr{B}$ there is a nonempty basis element $B^{\prime} \in \mathscr{B}^{\prime}$ such that $B^{\prime} \subset B$.- then it is unlikely that one can conclude (2'): one cannot use the original argument for showing the implication (2) $\Rightarrow$ (1) anymore.
- Can anyone give a counter-example for (2') $\Rightarrow$ (1)?
- When topologies are given by bases, one has a useful criterion in terms of the bases for determining whether one topology is finer than another:
- > Lemma 13.3. (Munkres's Topology p.81) Let $\mathscr{B}$ and $\mathscr{B}^{\prime}$ be bases for the topologies $\mathcal{T}$ and $\mathcal{T}^{\prime}$, respectively, on $X$. Then the following are equivalent:
- >
- > (1) $\mathcal{T}^{\prime}$ is finer than $\mathcal{T}$.
- > (2) For each $x \in X$ and each basis element $B \in \mathscr{B}$ containing $x$, there is a basis element $B^{\prime} \in \mathscr{B}^{\prime}$ such that $x \in B^{\prime} \subset B$.
- If one modifies condition (2) as:
- > (2') For each nonempty basis element $B \in \mathscr{B}$ there is a nonempty basis element $B^{\prime} \in \mathscr{B}^{\prime}$ such that $B^{\prime} \subset B$.
- then it is unlikely that one can conclude (2'): one cannot use the original argument for showing the implication (2) $\Rightarrow$ (1) anymore.
- Can anyone give a counter-example for (2') $\Rightarrow$ (1)?
#1: Initial revision
by
Snoopy
·
2023-02-03T15:06:07Z (almost 2 years ago)
Criterion in terms of the bases for determining whether one topology is finer than another
When topologies are given by bases, one has a useful criterion in terms of the bases for determining whether one topology is finer than another:
> Lemma 13.3. (Munkres's Topology p.81) Let $\mathscr{B}$ and $\mathscr{B}^{\prime}$ be bases for the topologies $\mathcal{T}$ and $\mathcal{T}^{\prime}$, respectively, on $X$. Then the following are equivalent:
>
> (1) $\mathcal{T}^{\prime}$ is finer than $\mathcal{T}$.
> (2) For each $x \in X$ and each basis element $B \in \mathscr{B}$ containing $x$, there is a basis element $B^{\prime} \in \mathscr{B}^{\prime}$ such that $x \in B^{\prime} \subset B$.
If one modifies condition (2) as:
> (2') For each basis element $B \in \mathscr{B}$ there is a nonempty basis element $B^{\prime} \in \mathscr{B}^{\prime}$ such that $B^{\prime} \subset B$.
then it is unlikely that one can conclude (2'): one cannot use the original argument for showing the implication (2) $\Rightarrow$ (1) anymore.
Can anyone give a counter-example for (2') $\Rightarrow$ (1)?