# Difference between finer and coarser topology

The terms "finer topology" and "coarser topology" describe a comparison between different topologies applied to the same set $ X $.

**Finer topology**

A finer topology on a set $ X $ is one that__has more open sets__than another topology on the same set.**Coarser topology**

A coarser topology, on the other hand,__has fewer open sets__compared to another topology on the same set $ X $. It's a simpler topology.

## A practical example

Let's consider the set \( X = \{a, b\} \) and apply different topologies to \( X \):

- The first topology \( \tau_1 \) is \( \{\varnothing, \{a, b\}\} \), the trivial topology, where only the empty set and the entire set are open.
- The second topology \( \tau_2 \) is \( \{\varnothing, \{a\}, \{a, b\}\} \).

The topology \( \tau_2 \) is finer than \( \tau_1 \) because it has more open sets, including \( \{a\} \), which is not open in \( \tau_1 \).

On the other hand, \( \tau_1 = \{\varnothing, \{a, b\}\} \) (the trivial topology) is coarser than \( \tau_2 \) because it has fewer open sets.

## Continuity in finer and coarser topology

If a function is continuous with respect to a coarser topology, it will also be continuous with respect to a finer topology (but the reverse is not necessarily true).

To verify the continuity of a function, we check that for every open set in the function’s image, its preimage is open in the domain.

So, when dealing with a finer topology that has more open sets, you have to check more sets for continuity.

Conversely, the coarser the topology, the easier it is to meet the continuity condition because there are fewer open sets to check.

In other words, **if a function is continuous with respect to a coarser topology, it will also be continuous with respect to a finer topology**, since if it satisfies the continuity conditions for the coarser topology, it will also do so for the same sets in the finer topology.

**The reverse, however, is not always true**. If a function is continuous with respect to a finer topology, it might not be continuous with respect to the coarser topology, as there are fewer open sets in the coarser topology, and the function might fail the continuity test for some preimages.

__Example__

Let’s revisit the example of the set \( X = \{a, b\} \), with two different topologies.

- Coarser topology: \( \tau_1 = \{\varnothing, \{a, b\}\} \), where only the empty set and the entire set \( X \) are open.
- Finer topology: \( \tau_2 = \{\varnothing, \{a\}, \{b\}, \{a, b\}\} \), where the subsets \( \{a\} \) and \( \{b\} \) are also open.

We now examine the continuity of a function \( f \) from \( X \) to a set \( Y \), where \( Y \) is any set.

$$ f: X \to Y $$

Define the function $ f $ as follows:

$$ f(a)=1 $$

$$ f(b)=1 $$

This function maps both \( a \) and \( b \) to 1, making it a constant function.

The function is continuous with respect to \( \tau_2 \) (the finer topology) because the preimages of \( \{a\} \) and \( \{a, b\} \) are open in \( X \).

- The preimage of \( f^{-1}(\{1\}) \), which is \( \{1\} \subseteq Y \), is \( \{a \} \cup \{ b \} = \{a, b\} \), since both \( a \) and \( b \) map to 1. The set \( \{a, b\} \) is open in \( X \) under the topology \( \tau_2 \), so the condition is satisfied.
- The preimage of the empty set \( \varnothing \) is still the empty set, \( f^{-1}(\varnothing) = \varnothing \), which by definition is open in \( X \) under any topology.

Thus, **\( f \) is continuous with respect to the finer topology \( \tau_2 \)**.

Since \( f \) is continuous with respect to \( \tau_2 \), it is automatically continuous with respect to \( \tau_1 \), because continuity in \( \tau_2 \) ensures that the preimage of \( \{a, b\} \) is open in \( X \).

This happens because the coarser topology \( \tau_1 \) only has two open sets to check: \( \varnothing \) and \( \{a, b\} \).

- The preimage \( f^{-1}(\{1\}) = \{a, b\} \) is open in \( \tau_1 \).
- The preimage \( f^{-1}(\varnothing) = \varnothing \) is also open in \( \tau_1 \).

Therefore, **\( f \) is also continuous with respect to the coarser topology \( \tau_1 \)**.

__Example 2__

Let’s again consider the set \( X = \{a, b\} \), with the topologies \( \tau_1 \) and \( \tau_2 \).

- Coarser topology: \( \tau_1 = \{\varnothing, \{a, b\}\} \).
- Finer topology: \( \tau_2 = \{\varnothing, \{a\}, \{b\}, \{a, b\}\} \).

Define a new function \( g : X \to Y \):

$$ g(a) = 1 $$

$$ g(b) = 2 $$

Now let’s check the continuity of \( g \) with respect to the finer topology \( \tau_2 \).

- The preimage \( f^{-1}( \varnothing ) = \varnothing \) is the empty set, which is open in \( \tau_2 \).
- The preimage \( f^{-1}(\{1,2\}) = \{a, b\} \) is open in \( \tau_2 \).
- The preimage \( f^{-1}(\{1\}) = \{a\} \) is open in \( \tau_2 \).
- The preimage \( f^{-1}(\{2\}) = \{b\} \) is open in \( \tau_2 \).

Since there are no other open sets to verify, **\( g \) is continuous with respect to the finer topology \( \tau_2 \)**.

Now let’s verify the continuity with respect to the coarser topology \( \tau_1 \):

- The preimage \( f^{-1}( \varnothing ) = \varnothing \) is the empty set, which is open in \( \tau_1 \).
- The preimage \( f^{-1}(\{1,2\}) = \{a,b\} \) is open in \( \tau_1 \).
- The preimage \( f^{-1}(\{1\}) = \{a\} \), but \( \{a\} \)
**is not an open set**in \( \tau_1 \).

This shows that **\( g \) is not continuous with respect to the coarser topology \( \tau_1 \)**.

In conclusion, the function \( g \) is continuous with respect to the finer topology \( \tau_2 \), but it is not continuous with respect to the coarser topology \( \tau_1 \).

And so on.