My topology text defines the "neighborhood" of a point as follows:Let p be a point in a topological space X .A subset N of X is a neighborhood of p iff N is a superset of an open set G containing p. Now I am stuck on the following problem:Let X be a cofinite topological space .Show that every neighborhood of a point p (element of x) is an open set .Now since in cofinite topology the closed sets are finite , so this should be true by definition , that is aclosed set could not contain an open set ...I just cannot see what we have to prove here ........


The definition is correct and also the statement that every neighborhood of a point in a space endowed with the cofinite topology is open.

Suppose $N$ is a neighborhood of $p$ and that $G$ is an open set such that $p\in G$ and $G\subseteq N$.

Then $X\setminus G\supseteq X\setminus N$, so $X\setminus N$ is finite. Hence $N$ is open.

  • $\begingroup$ Are we showing here that there exists one more larger open set $N$ in the topology, such that $G$ is a subset of it, and $G$ contains $x$ , hence making $N$ a nbd of $x$ as well as member of the topology ? I am a bit confused as to why $N$ should be open $\endgroup$ – The Doctor Apr 20 '18 at 6:23
  • $\begingroup$ @TheDoctor I'm proving that, if $N$ is a neighborhood of $p$, then $N$ is open. Since it is a neighborhood, there exists an open set $G$ with $p\in G\subseteq N$ (by definition of neighborhood). Now we check that $X\setminus N$ is finite, so that $N$ turns out to be open as well. $\endgroup$ – egreg Apr 20 '18 at 6:26
  • $\begingroup$ because $N'$ is finite, so $N$ is infinite and hence open. Am I right? $\endgroup$ – The Doctor Apr 20 '18 at 7:20
  • $\begingroup$ @TheDoctor Not really: $N'=X\setminus N$ finite ends the argument (the fact that $N$ is infinite is irrelevant and indeed false if $X$ is finite to begin with). $\endgroup$ – egreg Apr 20 '18 at 7:24
  • $\begingroup$ Thank you, I get it now $\endgroup$ – The Doctor Apr 20 '18 at 7:25

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