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I have very little knowledge of set theory and proof writing. This is a problem from “Introduction to Topology: Pure and Applied” by Colin Adams and Robert Franzosa.

DEFINITION

Let $X$ be a set and let $\mathcal{T}_1$ and $\mathcal{T}_2$ be two topologies on $X$. If $\mathcal{T}_1\subset\mathcal{T}_2$ then $\mathcal{T}_2$ is said to be finer than $\mathcal{T}_1$, and $\mathcal{T}_1$ is said to be coarser than $\mathcal{T}_2$. Furthermore, if $\mathcal{T}_2$ is finer than $\mathcal{T}_1$ but not equal to $\mathcal{T}_1$ then $\mathcal{T}_2$ is said to be strictly finer than $\mathcal{T}_1$. Strictly coarser is defined similarly.

In general, two topologies on a given set $X$ need not be comparable. Each may contain open sets that are not open sets in the other, and therefore neither topology would be finer than the other.

PROBLEM

Consider the following six topologies defined on $\mathbb{R}$: the trivial topology, the discrete topology, the finite complement topology, the standard topology, the lower limit topology, and the upper limit topology. Show how they compare to each other (finer, strictly finer, coarser, strictly coarser, noncomparable) and justify your claim.

THOUGHTS

  1. Trivial topology: define $\mathcal{T}_1=\{\varnothing,\mathbb{R}\}$
  2. Finite complement topology: define $\mathcal{T}_2$ as the topology whose open sets are the empty set and every set in $\mathbb{R}$ with a finite complement.
  3. Standard topology: define $\mathcal{T}_3$ as the topology whose open sets are the unions of sets of the type $(a,b)\subset\mathbb{R}$ such that $a<b$.
  4. Lower limit topology: define $\mathcal{T}_4$ as the topology whose open sets are the unions of sets of the type $[a,b)\subset\mathbb{R}$ such that $a<b$.
  5. Upper limit topology: define $\mathcal{T}_5$ as the topology whose open sets are the unions of sets of the type $(a,b]\subset\mathbb{R}$ such that $a<b$.
  6. Discrete topology: define $\mathcal{T}_6$ as the collection of all subsets of $\mathbb{R}$.

I believe I defined them correctly, but I'm not sure if they're in the right order from coarsest to finest, or if any are nomcomparable. Obviously $\mathcal{T}_1$ is strictly coarsest because it belongs to all other $\mathcal{T}_2-\mathcal{T}_6$. Likewise, $\mathcal{T}_6$ is strictly finest because $\mathcal{T}_1-\mathcal{T}_5$ belong to it. As always, I appreciate any help.

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    $\begingroup$ Terminology: $T_1$ does not belong to the others. It is a subset of the others. And the others do not belong to $T_6$. They are subsets of it. $\endgroup$ – DanielWainfleet Jun 15 at 2:11
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The $\mathcal{T}_1$ is strictly coarser than $\mathcal{T}_2$, which is strictly coarser than $\mathcal{T}_3$.

$\mathcal{T}_3$ is strictly coarser to both $\mathcal{T}_4$ and to $\mathcal{T}_5$ but $\mathcal{T}_4$ and $\mathcal{T}_5$ are not related, but incomparable, and of course all of these topologies are strictly coarser than $\mathcal{T}_6$, the discrete topology.

$[0,1) \in \mathcal{T}_4$ and $[0,1) \notin \mathcal{T}_5$ while $(0,1] \in \mathcal{T}_5$ and $(0,1] \notin \mathcal{T}_4$, hence the incomparability of these two topologies.

That $\mathcal{T}_3 \subsetneq \mathcal{T}_4$ is clear by $(a,b) = \bigcup \{[x,b): a < x < b\}$ and $[0,1) \notin \mathcal{T}_3$.

A similar argument can be made for $\mathcal{T}_3 \subsetneq \mathcal{T}_5$.

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  • $\begingroup$ If you wouldn't mind, could you show why T2 is strictly coarser than T3? $\endgroup$ – Math_Student_1 Jun 14 at 17:22
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    $\begingroup$ @Math_Student_1 T2 is the topology in which sets (other than R itself) are closed if and only if finite. In T3, all finite sets are closed but not all closed sets are finite. So for example the set $(0,1)$ is open in T3 but not in T2, because it complement is not finite. $\endgroup$ – MJD Jun 14 at 18:03

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