# Coarsest and finest topology

• shinobi20

#### shinobi20

Homework Statement
Verify that the “weakest” (coarsest) possible topology on a set ##X## is given by the trivial topology, where ∅ and ##X## represent the only open sets available, whereas the “strongest” (finest) topology is the discrete topology, where every subset is open.
Relevant Equations
1. ∅ ∈ {τ}, ##X## ∈ {τ};
2. the union (of an arbitrary number) of elements from {τ} is again in {τ};
3. the intersection of a finite number of elements from {τ} is again in {τ}.
I do not understand what is to verify here. The problem already defined what it means to be a trivial and discrete topology but it did not state what it means to be "weak" and "strong". I assume the problem wants me to connect "weak" with trivial topology and "strong" with discrete topology, but somehow the problem is not very clear to me or I just do not know how to connect them. Please guide me but do not give me the solution.

$T_1$ is weaker than $T_2$ iff $T_1 \subsetneq T_2$ (ie. every set which is open in $T_1$ is open in $T_2$, and there is at least one set which is open in $T_2$ but not in $T_1$.

$T_1$ is weaker than $T_2$ iff $T_1 \subsetneq T_2$ (ie. every set which is open in $T_1$ is open in $T_2$, and there is at least one set which is open in $T_2$ but not in $T_1$.
Did you mean ##T_2 \subsetneq T_1##? Clearly (I'll prove this but I'm just clarifying what you said), the trivial topology is a subset of the discrete topology but not the other way.

$T_1$ is weaker than $T_2$ iff $T_1 \subsetneq T_2$ (ie. every set which is open in $T_1$ is open in $T_2$, and there is at least one set which is open in $T_2$ but not in $T_1$.

The inclusion need not be strict, or at least this definition is not standard.

Given a set ##X##, there are multiple possible topologies on ##X##. Say we are given topologies ##\tau_1## and ##\tau_2## on ##X##. Then ##\tau_1 \subseteq \tau_2## means that ##\tau_1## is weaker than ##\tau_2## or equivalently ##\tau_2## is stronger than ##\tau_1##. This is just a definition.

Any set ##X## has a discrete topology ##2^X## (the power set of ##X##, i.e. all subsets of ##X## are open) and an indiscrete topology ##\{\emptyset, X\}##. It is quite trivial to see that these are topologies.

The question asks you to show that if ##\tau## is a topology on ##X##, then ##\{\emptyset, X\}\subseteq \tau \subseteq 2^X## and half a second of thought shows you that this is completely trivial.

I guess your confusion comes from the fact that this "problem" is so trivial that it is not clear what exactly there is to verify.

Did you mean ##T_2 \subsetneq T_1##? Clearly (I'll prove this but I'm just clarifying what you said), the trivial topology is a subset of the discrete topology but not the other way.

I meant what I said: $\{\emptyset, X\} \subsetneq 2^X$.

The inclusion need not be strict, or at least this definition is not standard.

I'm not sure it makes sense to say that a topology is weaker than itself.

I meant what I said: $\{\emptyset, X\} \subsetneq 2^X$.

I'm not sure it makes sense to say that a topology is weaker than itself.

Of course that makes sense! Two topologies coincide when one is both weaker and stronger than the other. And take ##X=\{0\}##. The trivial topologies coincide so your strict inclusion is false.

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$T_1$ is weaker than $T_2$ iff $T_1 \subsetneq T_2$ (ie. every set which is open in $T_1$ is open in $T_2$, and there is at least one set which is open in $T_2$ but not in $T_1$.
This is contradictory to the statement of @Math_QED, which is

Given a set X, there are multiple possible topologies on X. Say we are given topologies τ1 and τ2 on X. Then τ1⊆τ2 means that τ1 is weaker than τ2 or equivalently τ2 is stronger than τ1. This is just a definition.

The question asks you to show that if τ is a topology on X, then {∅,X}⊆τ⊆2X and half a second of thought shows you that this is completely trivial.
If this is the question, then a simple proof should go as,

Proof. From the definition that in order to introduce a topology ##τ## on a set ##X##, the sets ##∅## and ##X## must belong to all possible topologies ##τ## of ##X##, then it is guaranteed that ##\{∅, X\}## is a subset of all ##τ## and since ##2^X## contains all subsets of ##X## it is guaranteed that all possible topologies ##τ## is a subset of ##2^X##.

This implies that ##\{∅, X\} ⊆ τ ⊆2^X##. So the trivial topology is the weakest and the discrete topology is the strongest.

• member 587159
If this is the question, then a simple proof should go as,

Proof. From the definition that in order to introduce a topology ##τ## on a set ##X##, the sets ##∅## and ##X## must belong to all possible topologies ##τ## of ##X##, then it is guaranteed that ##\{∅, X\}## is a subset of all ##τ## and since ##2^X## contains all subsets of ##X## it is guaranteed that all possible topologies ##τ## is a subset of ##2^X##.

This implies that ##\{∅, X\} ⊆ τ ⊆2^X##. So the trivial topology is the weakest and the discrete topology is the strongest.

Exactly, in other words, if we consider the set of all topologies on ##X## and we partially order it via the inclusion relation, then this set has both a minimum and a maximum.

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Exactly, in other words, if we consider the set of all topologies on ##X## and we partially order it via the inclusion relation, then this set has both a minimum and a minimum.
I understand everything now. Thanks @Math_QED !

• member 587159