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

[Maths Lover]

My question is:

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

Show that: f 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.

 

[micromass]

What definition of continuity would you use?

 

[Maths Lover]

What definition of continuity would you use?

ِA function f:X \rightarrow Y between two topological spaces X and Y is continuous if the preimage of any open set of Y is an open set of X.

 

[micromass]

ِA function f:X \rightarrow Y between two topological spaces X and Y is continuous if the preimage of any open set of Y is an open set of X.

Yes. Do you perhaps know of others characterizations of continuity that would be more handy in this case?

 

[Maths Lover]

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 x intersects the set itself.

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

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 :)