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.
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
.
We cannot pick an element from an empty set, so h does not exist.
Now consider an empty function g such that
.
By definition of a function, g can be described as,
"g is a relation F such that for each x in the domain , there is a unique y in satisfying xFy."
The above is vacuously true since there are not any x in the . 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 . 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 is continuous.