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Suppose $X$ is a finite set with more than $5$ elements. Which of the following is true.

  1. There is a topology on $X$ which is $T_3$.
  2. There is a topology on $X$ which is $T_2$ but not $T_3$.
  3. There is a topology on $X$ which is $T_1$ but not $T_2$.
  4. There is no topology on $X$ which is $T_1$

First how to determine which topology is there? Since it is a finite set, so the topologies are a subset of $P(X)$, power set of $X$. We can consider the discrete topology. What are other kind of topologies we can consider so it match with the given options. Please help me with this. Thanks.

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    This is a weird question...the "more than 5 elements" condition is totally irrelevant. – 2017-01-11
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    @eric that's what i also thought. It is there just to add confusion. – 2017-01-11
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    @Eric: It may also be because the topologies on sets with 4 or fewer elements have previously been completely classified, and the reader ia being asked to generalize or break the generalization for other finite cardinalities. – 2017-01-11
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    Does $T_3$ imply $T_1$ in your definition. Or is it only being able to separate a closed set from a point? – 2017-01-11
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    @HennoBrandsma $T_3$ is regular, isn't it? Separate a closed set from a point. – 2017-01-11

2 Answers 2

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Hint: Since $X$ is finite, you should try to figure out what a $T_1$ topology on $X$ can look like. This should fully allow you to evaluate all four of the proposed statements.

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    Ok so if i am considering a discrete topology then as it satisfy all separation axioms so option 1 is true. Right? – 2017-01-11
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    Yes, the first statement is true. What about the others? – 2017-01-11
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    I am only considering discrete topology, I don't know any other topology on a finite set which is $T_1$. – 2017-01-11
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    In fact, one can prove that if a topology on a finite set is $T_1,$ then it *must* be discrete. – 2017-01-11
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    Given that result (which I leave to you to prove), can you determine if the other three statements are true, false, or undetermined? – 2017-01-11
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    Oh I didn't know that. – 2017-01-11
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    Ok provided I can prove the result, we can readily get rid of option 4,3,2. – 2017-01-11
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    Nicely concluded! Good luck with the proof, and let me know if you get stuck. – 2017-01-11
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    Ok thanks for the hint. – 2017-01-11
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A $X$ is a finite $T_1$ space, every singleton set $\{x\}$ is closed. As closed sets are closed under finite unions, all subsets of $X$ are closed and so all subsets of $X$ are also open. Hence $X$ is discrete.

Nr 4. Is false: the discrete topology is $T_1$. If $X$ is finite and $T_2$ or $T_1$, it will be at least $T_1$ so discrete and $T_3,T_2, T_1$ as well. So 2 and 3 are false. 1 is true as the discrete topology again is an example. The number 5 is irrelevant, finiteness is all that matters.