Topologies

Geometry Level 3

Which of the following collections $\mathcal T$ of subsets of $\mathbb R$ is a topology on $\mathbb R$ ?

I. $\mathcal T =$ the empty set, $\mathbb R$ , and all intervals of the form $[a,\infty)$ for any $a \in \mathbb R$
II. $\mathcal T =$ the empty set, plus all subsets $Y \subseteq \mathbb R$ such that the complement ${\mathbb R} \setminus Y$ is finite
III. $\mathcal T =$ the empty set, plus all infinite subsets $Y \subseteq \mathbb R$

III only I only I and II I and III II and III II only

1 solution

Patrick Corn
Jun 15, 2016

I is not a topology, since the union of the intervals $[1/n,\infty)$ for $n=1,2,3,\ldots$ is $(0,\infty),$ which is not one of the sets in $\mathcal T.$

II is a topology on any set, not just $\mathbb R.$ It is known as the cofinite topology .

III is not a topology, as the intersection of two infinite sets does not need to be infinite (e.g. $[0,1]$ and $[1,2]$ are in $\mathcal T$ but their intersection $\{1\}$ is not).

So the answer is "II only."

A Question about the cofinite topology. If we have a collection A defined: (-inf,...,n) U (n,...,inf) for every n in Natural set. then, the Union U(A) is infinite ==> U(A) not in cofinite topology. according to the definitions:The union of a collection of sets in T is in T where did I get it wrong?

אסף שובל - 2 years, 6 months ago

I don't understand. The union of the sets in your collection (indeed the union of any two distinct sets in your collection) is all of $\mathbb R,$ so it's open in the cofinite topology because its complement, the empty set, is finite.

Patrick Corn - 2 years, 6 months ago

Collection A is a subset of R, complement R \ A in your case is {n}, which is finite.

Nirav Baid - 2 years, 3 months ago

Topologies

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