Function space is weakly Hausdorff?
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Let $X^I$ denote the set of continuous maps $f:I rightarrow X$, $X$ weakly Hausdorff, $I$ unit interval. Given the compact open topology, is the space weakly Hausdorff?
To define weakly Hausdorff: for all compact hausdorf spaces, $K$, and for all continuous functions $t:K rightarrow X^I$: $t[K]$ is closed in $X^I$.
general-topology algebraic-topology
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Let $X^I$ denote the set of continuous maps $f:I rightarrow X$, $X$ weakly Hausdorff, $I$ unit interval. Given the compact open topology, is the space weakly Hausdorff?
To define weakly Hausdorff: for all compact hausdorf spaces, $K$, and for all continuous functions $t:K rightarrow X^I$: $t[K]$ is closed in $X^I$.
general-topology algebraic-topology
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X^I$ denote the set of continuous maps $f:I rightarrow X$, $X$ weakly Hausdorff, $I$ unit interval. Given the compact open topology, is the space weakly Hausdorff?
To define weakly Hausdorff: for all compact hausdorf spaces, $K$, and for all continuous functions $t:K rightarrow X^I$: $t[K]$ is closed in $X^I$.
general-topology algebraic-topology
Let $X^I$ denote the set of continuous maps $f:I rightarrow X$, $X$ weakly Hausdorff, $I$ unit interval. Given the compact open topology, is the space weakly Hausdorff?
To define weakly Hausdorff: for all compact hausdorf spaces, $K$, and for all continuous functions $t:K rightarrow X^I$: $t[K]$ is closed in $X^I$.
general-topology algebraic-topology
edited Jul 15 at 16:57
Henno Brandsma
91.6k342100
91.6k342100
asked Jul 15 at 10:09
Cyryl L.
1,7112821
1,7112821
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