# Prove that $f: X \rightarrow Y$ is a continuous function.

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.

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What definition of continuity would you use?

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 $X$.

ِ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?

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 :)