# Finer/coarser topologies

1. Dec 6, 2011

### Ted123

1. The problem statement, all variables and given/known data

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 $(X,\tau_2)$.

3. 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?

2. Dec 6, 2011

### spamiam

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.

3. Dec 7, 2011

### Ted123

$\tau_2 \subseteq \tau_1$ 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

4. Dec 7, 2011

### spamiam

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.

5. Dec 7, 2011

### Ted123

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?)

6. Dec 7, 2011

### spamiam

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.