Coarsest and finest topology

[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.

 

[pasmith]

$T_1$ 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 $T_1$.

 

[shinobi20]

$T_1$ 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.

 

[member 587159]

$T_1$ 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.

 

[member 587159]

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.

 

[pasmith]

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.

 

[member 587159]

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.

 

[shinobi20]

$T_1$ 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.

 

[shinobi20]

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 !