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Math Help - trivial topology in domain

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by Andreamet View Post
    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.
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  3. #3
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    Quote Originally Posted by robeuler View Post
    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.
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  4. #4
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    Quote Originally Posted by memory loss View Post
    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.
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