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removed deprecated tag 'topology'
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Ricardo Andrade
Ricardo Andrade
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added link, improved grammar
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edit approved Feb 2, 2014 at 11:47
Rasmus
edit approved Feb 2, 2014 at 11:47
Rasmus
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Every topological space gives rise to a pseudotopological spacepseudotopological space. Conversely, if $X$ is a pseudotopological space then we can define a topology on $X$ such that every filter converging to $x$ in the original pseudotopology also converges in this topology.

Let $(X,T)$ be a topological space. Let $F$ a subspace of $(X,T)$ considered as a pseudotopological space. My question is that whether $F$ necessarily is a subspace of the topological space $(X,T)$ or not.

Every topological space gives rise to a pseudotopological space. Conversely, if $X$ is a pseudotopological space then we can define a topology on $X$ such that every filter converging to $x$ in the original pseudotopology also converges in this topology.

Let $(X,T)$ be a topological space. Let $F$ a subspace of $(X,T)$ considered as pseudotopological space. My question is that whether $F$ is a subspace of topological space $(X,T)$ or not.

Every topological space gives rise to a pseudotopological space. Conversely, if $X$ is a pseudotopological space then we can define a topology on $X$ such that every filter converging to $x$ in the original pseudotopology also converges in this topology.

Let $(X,T)$ be a topological space. Let $F$ a subspace of $(X,T)$ considered as a pseudotopological space. My question is whether $F$ necessarily is a subspace of the topological space $(X,T)$ or not.

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Hina
Hina
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subspace in pseudotopological space

Every topological space gives rise to a pseudotopological space. Conversely, if $X$ is a pseudotopological space then we can define a topology on $X$ such that every filter converging to $x$ in the original pseudotopology also converges in this topology.

Let $(X,T)$ be a topological space. Let $F$ a subspace of $(X,T)$ considered as pseudotopological space. My question is that whether $F$ is a subspace of topological space $(X,T)$ or not.