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

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- Apr 3rd 2009, 10:21 AMAndreamettrivial 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 - Apr 3rd 2009, 10:55 AMrobeuler
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. - Apr 7th 2009, 01:06 AMmemory 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.

- Apr 7th 2009, 02:54 AMaliceinwonderland
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

$\displaystyle h : A \rightarrow \emptyset$.

We cannot pick an element from an empty set, so h does not exist.

Now consider an empty function g such that

$\displaystyle g : \emptyset \rightarrow \emptyset$.

By definition of a function, g can be described as,

"g is a relation F such that for each x in the domain $\displaystyle \emptyset$, there is a unique y in $\displaystyle \emptyset$ satisfying xFy."

The above is vacuously true since there are not any x in the $\displaystyle \emptyset$. 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 $\displaystyle f:X \rightarrow Y$. 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 $\displaystyle h:X \rightarrow Y$ is continuous.