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Prove that [itex]f: X \rightarrow Y[/itex] is a continuous function.

  1. Aug 15, 2014 #1
    My question is:

    Let [itex]f:\bigcup_{\alpha}A_{\alpha} \rightarrow Y[/itex] be a function between the topological spaces Y and [itex]X=\bigcup_{\alpha}A_{\alpha}[/itex]. Suppose that [itex]f|A_{\alpha}[/itex] is a continuous function for every [itex]\alpha[/itex] and that [itex]{A_{\alpha}}[/itex] is locally finite collection. Suppose that [itex]A_{\alpha}[/itex] is closed for every [itex]\alpha[/itex].

    Show that: [itex]f[/itex] is continuous.

    Any hints?

    I'm stuck with this problem for some days. Some gave me answers on mathematics stackexchange. but it didn't make much sense.
     
  2. jcsd
  3. Aug 15, 2014 #2

    micromass

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    What definition of continuity would you use?
     
  4. Aug 15, 2014 #3
    ِA function [itex]f:X \rightarrow Y[/itex] between two topological spaces [itex]X[/itex] and [itex]Y[/itex] is continuous if the preimage of any open set of [itex]Y[/itex] is an open set of [itex]X[/itex].
     
  5. Aug 15, 2014 #4

    micromass

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    Yes. Do you perhaps know of others characterizations of continuity that would be more handy in this case?
     
  6. Aug 16, 2014 #5
    Sure, I know that a function is continuous if the preimage of an closed set is closed. and a set is closed iff its closure is the set itself. I lately heared that an element x is in the closure of a set if every neighborhood of [itex]x[/itex] intersects the set itself.

    a function is continuous iff for every [itex]x[/itex] in the domain. the preimage of a neighborhood [itex]U[/itex] of [itex]f(x)[/itex] is a neighborhood of [itex]x[/itex].

    Note: I got an answer for the question here (Mathematics StackExchange )but I still concerned in knowing different ideas for the problem as it puzzeled me for several days :)
     
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