Proving Continuity in Stronger and Weaker Topologies

[Ted123]

Homework Statement

Suppose \tau_1 and \tau_2 are 2 topologies on a set X and that \tau_2 \subseteq \tau_1. We say that \tau_1 is stronger/finer than \tau_2 and that \tau_2 is weaker/coarser than \tau_1.

Show, directly from the definitions, that if:

(a) A \subseteq X is closed in (X,\tau_2) then A is closed in (X,\tau_1) ;

(b) (Y,\tau_Y) is another topological space and f is a continuous map from (Y,\tau_Y) to (X,\tau_1) then f is continuous from (Y,\tau_Y) to (X,\tau_2).

The Attempt at a Solution

For (a), if A \subseteq X is closed in (X,\tau_2) then, by definition, \partial A \subseteq A, but this is precisely the definition of A being closed in (X,\tau_1) (the definition is dependent on the set, not the topology).

Is this OK - is there a better way to show it?

For (b), f:(Y,\tau_Y) \to (X,\tau_1) is continuous if for every open set A\subseteq X,\; f^*(A) is open in Y. Again, isn't this just the definition of f:(Y,\tau_Y) \to (X,\tau_2) being continuous?

 

[spamiam]

Homework Statement

Suppose \tau_1 and \tau_2 are 2 topologies on a set X and that \tau_2 \subseteq \tau_1. We say that \tau_1 is stronger/finer than \tau_2 and that \tau_2 is weaker/coarser than \tau_1.

Show, directly from the definitions, that if:

(a) A \subseteq X is closed in (X,\tau_2) then A is closed in (X,\tau_1) ;

(b) (Y,\tau_Y) is another topological space and f is a continuous map from (Y,\tau_Y) to (X,\tau_1) then f is continuous from (Y,\tau_Y) to (X,\tau_2).

The Attempt at a Solution

For (a), if A \subseteq X is closed in (X,\tau_2) then, by definition, \partial A \subseteq A, but this is precisely the definition of A being closed in (X,\tau_1) (the definition is dependent on the set, not the topology).

This last comment isn't true, at least not the way I understand it. What is your definition of the boundary of A? It must depend on the topology of the space in some way. Where have you used the fact that \tau_2 \subseteq \tau_1?

Is this OK - is there a better way to show it?

The definition of closed I learned for a closed set is F is closed iff X \setminus F is open. You could try starting there.

For (b), f:(Y,\tau_Y) \to (X,\tau_1) is continuous if for every open set A\subseteq X,\; f^*(A) is open in Y. Again, isn't this just the definition of f:(Y,\tau_Y) \to (X,\tau_2) being continuous?

Again, where have you used the fact that \tau_2 \subseteq \tau_1? Maybe you're just omitting these because they seem clear, but I think it's important to mention where you use each piece of information.

 

[Ted123]

This last comment isn't true, at least not the way I understand it. What is your definition of the boundary of A? It must depend on the topology of the space in some way. Where have you used the fact that \tau_2 \subseteq \tau_1?

The definition of closed I learned for a closed set is F is closed iff X \setminus F is open. You could try starting there.


Again, where have you used the fact that \tau_2 \subseteq \tau_1? Maybe you're just omitting these because they seem clear, but I think it's important to mention where you use each piece of information.

\tau_2 \subseteq \tau_1 means every \tau_2-open set is \tau_1-open so for (a):

A is closed in (X,\tau_2) \implies A^c is \tau_2-open in X

\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\, \implies A^c is \tau_1-open in X (since \tau_2 \subseteq \tau_1)

\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\, \implies A is closed in (X,\tau_1)

Does that look better?

However is every \tau_1-open set \tau_2-open? As for (b):

f:(Y,\tau_Y) \to (X,\tau_1) is continuous \implies for every \tau_1-open set A \subseteq X,\; f^*(A) is \tau_Y-open in Y

Does this imply the following?

\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\, \implies for every \tau_2-open set A \subseteq X,\; f^*(A) is \tau_Y-open in Y (since \tau_2 \subseteq \tau_1??)


\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\, \implies f:(Y,\tau_Y) \to (X,\tau_2) is continuous

 

[spamiam]

\tau_2 \subseteq \tau_1 means every \tau_2-open set is \tau_1-open so for (a):

A is closed in (X,\tau_2) \implies A^c is \tau_2-open in X

\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\, \implies A^c is \tau_1-open in X (since \tau_2 \subseteq \tau_1)

\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\, \implies A is closed in (X,\tau_1)

Does that look better?

Looks good.

However is every \tau_1-open set \tau_2-open?

This would only be true if \tau_1 \subseteq \tau_2, in which case the 2 topologies are equal.

As for (b):

f:(Y,\tau_Y) \to (X,\tau_1) is continuous \implies for every \tau_1-open set A \subseteq X,\; f^*(A) is \tau_Y-open in Y

Does this imply the following?

\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\, \implies for every \tau_2-open set A \subseteq X,\; f^*(A) is \tau_Y-open in Y (since \tau_2 \subseteq \tau_1??)


\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\, \implies f:(Y,\tau_Y) \to (X,\tau_2) is continuous

This is fine, but you seem unsure, so maybe it would be better to start with an arbitrary open set O in (X, \tau_2) and prove that f^{-1}(O) is open in Y.

 

[Ted123]

Looks good.

This would only be true if \tau_1 \subseteq \tau_2, in which case the 2 topologies are equal.



This is fine, but you seem unsure, so maybe it would be better to start with an arbitrary open set O in (X, \tau_2) and prove that f^{-1}(O) is open in Y.

The thing I was unsure about was every \tau_1-open set... implying every \tau_2-open set...

(as you say, wouldn't this only be true if \tau_1 \subseteq \tau_2, in which case the 2 topologies are equal?)

 

[spamiam]

The thing I was unsure about was every \tau_1-open set... implying every \tau_2-open set...

(as you say, wouldn't this only be true if \tau_1 \subseteq \tau_2, in which case the 2 topologies are equal?)

You've got it backwards. You don't need every set in \tau_1 to be in \tau_2 to prove the statement in part b). As I said before, try starting with an arbitrary open set O in (X, \tau_2) and prove that f^{-1}(O) is open in Y. Writing it out is the best way to make this clear.