# Thread: trivial topology in domain

1. ## trivial topology in domain

Correction to title: Trivial topology in RANGE

Let Y ave the trivial topology and X be an arbitrary topological space. Show that every function f: X->Y is continuous

2. Originally Posted by Andreamet
Correction to title: Trivial topology in RANGE

Let Y ave the trivial topology and X be an arbitrary topological space. Show that every function f: X->Y is continuous
In Y the only closed sets are Y and the empty set. All others are open.

Do you know the open/closed set formulation of continuity? It says that if the preimage of open (closed) sets are open (closed), then f is continuous.

If you can show that the preimages of Y and the empty set are closed, then you are done.

3. Originally Posted by robeuler
In Y the only closed sets are Y and the empty set. All others are open.

Do you know the open/closed set formulation of continuity? It says that if the preimage of open (closed) sets are open (closed), then f is continuous.

If you can show that the preimages of Y and the empty set are closed, then you are done.
If A is open set in Y then it must be empty set or all of Y. Preimage of empty set is empty set and preimage of all of Y is entire domain space X right? I think that is all because there are no more open sets in Y to consider and this must be true of any function from X to Y.

4. Originally Posted by memory loss
If A is open set in Y then it must be empty set or all of Y. Preimage of empty set is empty set and preimage of all of Y is entire domain space X right? I think that is all because there are no more open sets in Y to consider and this must be true of any function from X to Y.
Right.

For (a),

The reason why the preimage of the empty set under f is the empty set involves a little bit of set theory.

The definition of a function is
"A function is a relation F such that for each x in dom F there is only one y such that xFy".

Consider a nonempty set A such that
$h : A \rightarrow \emptyset$.
We cannot pick an element from an empty set, so h does not exist.

Now consider an empty function g such that
$g : \emptyset \rightarrow \emptyset$.

By definition of a function, g can be described as,
"g is a relation F such that for each x in the domain $\emptyset$, there is a unique y in $\emptyset$ satisfying xFy."

The above is vacuously true since there are not any x in the $\emptyset$. So g exists.

Now going back to a function f between topological spaces, the preimage of an empty set in Y is uniquely determined, which is an empty set in the topological space X. No other subset of X except the empty set can be mapped to the empty set in the topological space Y. The empty set is open in every topological space.

For (b), the preimage of Y is X, since f is defined as $f:X \rightarrow Y$. We know that X is open in the topological space X.

By definition of a continuous function in topology, f is continuous.

A similar example is
If X has the discrete topology, then any map $h:X \rightarrow Y$ is continuous.