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I want to follow the discussion from here concerning about the strength of the separation "all retract subspaces are closed".

(A retract subspace of a topological space $X$ is a subspace $A$ where there exists continuous $f: X\to A$ such that $f|_A = \mathrm{id}_A$.)

Write $\mathrm{KC}$ as "all compact subsets are closed", and $\mathrm{RC}$ as "all retract subspaces are closed". We have $T_2\Rightarrow \mathrm{KC}$, $T_2\Rightarrow \mathrm{TC}$ and neither of the reversed implications. The cocountable topology over $\mathbb{R}$ is a $\mathrm{KC}$ example not $\mathrm{RC}$: $f(x) = |x|$ is a continuous function from $\mathbb{R}$ with cocountable topology to its subspace $[0,+\infty)$ since the preimages of countable sets are countable, but $[0,+\infty)$ is certainly not closed.

So I would like to ask if $\mathrm{RC}$ implies $\mathrm{KC}$, because the only $\mathrm{RC}$ non-Hausdorff spaces I know are those compact $\mathrm{KC}$ spaces (note that compact $\mathrm{KC}$ implies $\mathrm{RC}$). Any help appreciated.

I want to follow the discussion from here concerning about the strength of the separation "all retract subspaces are closed".

(A retract subspace of a topological space $X$ is a subspace $A$ where there exists continuous $f: X\to A$ such that $f|_A = \mathrm{id}_A$.)

Write $\mathrm{KC}$ as "all compact subsets are closed", and $\mathrm{RC}$ as "all retract subspaces are closed". We have $T_2\Rightarrow \mathrm{KC}$, $T_2\Rightarrow \mathrm{RC}$ and neither of the reversed implications. The cocountable topology over $\mathbb{R}$ is a $\mathrm{KC}$ example not $\mathrm{RC}$: $f(x) = |x|$ is a continuous function from $\mathbb{R}$ with cocountable topology to its subspace $[0,+\infty)$ since the preimages of countable sets are countable, but $[0,+\infty)$ is certainly not closed.

So I would like to ask if $\mathrm{RC}$ implies $\mathrm{KC}$, because the only $\mathrm{RC}$ non-Hausdorff spaces I know are those compact $\mathrm{KC}$ spaces (note that compact $\mathrm{KC}$ implies $\mathrm{RC}$). Any help appreciated.

I want to follow the discussion from here concerning about the strength of the separation "all retract subspaces are closed".

(A retract subspace of a topological space $X$ is a subspace $A$ where there exists continuous $f: X\to A$ such that $f|_A = \mathrm{id}_A$.)

Write $\mathrm{KC}$ as "all compact subsets are closed", and $\mathrm{RC}$ as "all retract subspaces are closed". We have $T_2\Rightarrow \mathrm{KC}$, $T_2\Rightarrow \mathrm{TC}$ and neither the reversed implications. The cocountable topology over $\mathbb{R}$ is a $\mathrm{KC}$ example not $\mathrm{RC}$: $f(x) = |x|$ is a continuous function from $\mathbb{R}$ with cocountable topology to its subspace $[0,+\infty)$ since the preimages of countable sets are countable, but $[0,+\infty)$ is certainly not closed.

So I would like to ask if $\mathrm{RC}$ implies $\mathrm{KC}$, because the only $\mathrm{RC}$ non-Hausdorff spaces I know are those compact $\mathrm{KC}$ spaces (note that compact $\mathrm{KC}$ implies $\mathrm{RC}$). Any help appreciated.

I want to follow the discussion from here concerning about the strength of the separation "all retract subspaces are closed".

(A retract subspace of a topological space $X$ is a subspace $A$ where there exists continuous $f: X\to A$ such that $f|_A = \mathrm{id}_A$.)

Write $\mathrm{KC}$ as "all compact subsets are closed", and $\mathrm{RC}$ as "all retract subspaces are closed". We have $T_2\Rightarrow \mathrm{KC}$, $T_2\Rightarrow \mathrm{TC}$ and neither of the reversed implications. The cocountable topology over $\mathbb{R}$ is a $\mathrm{KC}$ example not $\mathrm{RC}$: $f(x) = |x|$ is a continuous function from $\mathbb{R}$ with cocountable topology to its subspace $[0,+\infty)$ since the preimages of countable sets are countable, but $[0,+\infty)$ is certainly not closed.

So I would like to ask if $\mathrm{RC}$ implies $\mathrm{KC}$, because the only $\mathrm{RC}$ non-Hausdorff spaces I know are those compact $\mathrm{KC}$ spaces (note that compact $\mathrm{KC}$ implies $\mathrm{RC}$). Any help appreciated.

I want to follow the discussion from here concerning about the strength of the separation "all retract subspaces are closed".

(A retract subspace of a topological space $X$ is a subspace $A$ where there exists continuous $f: X\to A$ such that $f|_A = \mathrm{id}_A$.)

Write $\mathrm{KC}$ as "all compact subsets are closed", and $\mathrm{RC}$ as "all retract subspaces are closed". We have $T_2\Rightarrow \mathrm{KC}$, $T_2\Rightarrow \mathrm{TC}$ and neither the converse implications. The cocountable topology over $\mathbb{R}$ is a $\mathrm{KC}$ example not $\mathrm{RC}$: $f(x) = |x|$ is a continuous function from $\mathbb{R}$ with cocountable topology to its subspace $[0,+\infty)$ since the preimages of countable sets are countable, but $[0,+\infty)$ is certainly not closed.

So I would like to ask if $\mathrm{RC}$ implies $\mathrm{KC}$, because the only $\mathrm{RC}$ non-Hausdorff spaces I know are those compact $\mathrm{KC}$ spaces (note that compact $\mathrm{KC}$ implies $\mathrm{RC}$). Any help appreciated.

I want to follow the discussion from here concerning about the strength of the separation "all retract subspaces are closed".

(A retract subspace of a topological space $X$ is a subspace $A$ where there exists continuous $f: X\to A$ such that $f|_A = \mathrm{id}_A$.)

Write $\mathrm{KC}$ as "all compact subsets are closed", and $\mathrm{RC}$ as "all retract subspaces are closed". We have $T_2\Rightarrow \mathrm{KC}$, $T_2\Rightarrow \mathrm{TC}$ and neither the reversed implications. The cocountable topology over $\mathbb{R}$ is a $\mathrm{KC}$ example not $\mathrm{RC}$: $f(x) = |x|$ is a continuous function from $\mathbb{R}$ with cocountable topology to its subspace $[0,+\infty)$ since the preimages of countable sets are countable, but $[0,+\infty)$ is certainly not closed.

So I would like to ask if $\mathrm{RC}$ implies $\mathrm{KC}$, because the only $\mathrm{RC}$ non-Hausdorff spaces I know are those compact $\mathrm{KC}$ spaces (note that compact $\mathrm{KC}$ implies $\mathrm{RC}$). Any help appreciated.