[Math] Weak topology generated by family of functions is weakest wrt which each of the given functions is continuous

general-topologyreal-analysis

Let $\mathcal{F}=\{ f_\alpha:X \rightarrow X_\alpha, \alpha \in A\}$ be a family of functions from X onto topological spaces $X_\alpha$. The weak topology for X generated by functions $f_\alpha$ is the topology determined by the subbasis of open sets $\{ f^{-1}_\alpha (O_\alpha): \alpha \in \mathcal{A}, O_\alpha$ open in $X_\alpha\}$.

Why would the weak topology defined above be the weakest topology with respect to each of which of the given function is continuous?

Suppose we have a family of constant functions (they are continuous), then the topology generated is just the trivial topology, wouldn't it in this case be weaker than the weak topology also?

Best Answer

If we want our functions $f_\alpha$ to be continuous, then the preimage of any open set has to be open (by definition), therefore, all the sets $f^{-1}(O_\alpha)$ have to be open.

On the other hand, this family of sets satisfies the definition of being the basis of a topology (should be quite easy to verify).

Therefore, this topology in minimal (i.e. the weakest) among topologies which make the functions $f_\alpha$ continuous - it simply has nothing but necessary sets for continuity.