A topological space (X,T) is said to be quasi compact if every open covering has a finite refinement. Let f be a fuction from (X,T) to another topological space (Y,T') then f is said to be quasi compact if inverse image of every quasi compact open subset of Y is a quasi compact subset of X.

Recently I saw another definition of quasi compactness in the research paper On Door Spaces by Julian Dontchev which says that a function f from a topological space (X,T) to another topological space (Y,T') is called quasi compact if it satisfies the following condition:

Let U be an open subset of X such that inverse image of f(U) is U itself then f(U) is open in Y. My doubt is whether this two definitions are equivalent.If yes kindly post the proof .


1 Answer 1


These certainly are not equivalent. For instance, let both spaces be $\mathbb{R}$ with the usual topology. The only quasicompact open subset of $\mathbb{R}$, so trivially any map $f:\mathbb{R}\to\mathbb{R}$ is quasicompact by the first definition. But it's certainly not true that every map $f:\mathbb{R}\to\mathbb{R}$ is quasicompact by the second definition. For instance, $f(x)=x^2$ fails the definition for $U=\mathbb{R}$.


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