0

Let $\{A_i\}_{i\in I}$ be a collection of subsets of a topological space $X$, such that $X = \bigcup_{i \in I} A_i$. Let $f:X \rightarrow Y$ be a map of topological spaces such that the restriction of $f$ to each $A_i$ (equipped with the subspace topology induced from $X$) is continuous.

Show that if each $A_i$ is closed, and $I$ is finite, then $f$ is continuous.

I understand that this can be solved by the pasting lemma, but I don't see how I can do it using both hypotheses that $A_i$ is closed and $I$ finite.

A. Attia
  • 105
  • 5

1 Answers1

1

Let $F\subseteq Y$ be closed.

Then $f^{-1}(F)\cap A_i$ is closed in $A_i$ for every $i\in I$ because $f$ restricted to $A_i$ is continuous for every $i\in I$.

That means that we can write $f^{-1}(F)\cap A_i=A_i\cap G$ for some closed $G\subseteq X$.

But then $A_i\cap G$ is - as an intersection of closed sets - also closed (here it is used that $A_i$ is closed).

Then $f^{-1}(F)=\bigcup_{i\in I}(f^{-1}(F)\cap A_i)$ is a finite union of closed sets, hence is closed (here it is used that $I$ is finite).

This proves that $f$ is continuous.

drhab
  • 151,093