Proposition 8

Let $X$ be a set, $(X_\lambda)_{\lambda \in L}$ a family of subsets of $X$. Suppose each $X_\lambda$ carries a topology $\mathfrak{T}_\lambda$ such that, for each pair of indices $(\lambda, \mu)$: 1) $X_\lambda \cap X_\mu$ is open (resp. closed) in each of the topologies $\mathfrak{T}_\lambda, \mathfrak{T}_\mu$. 2) The topologies induced on $X_\lambda \cap X_\mu$ by $\mathfrak{T}_\lambda$ and $\mathfrak{T}_\mu$ coincide. Let $\mathfrak{T}$ be the finest topology on $X$ for which the injections $j_\lambda : X_\lambda \to X$ are continuous. Then, for each $\lambda \in L$, $X_\lambda$ is open (resp. closed) in $X$ in the topology $\mathfrak{T}$, and the topology induced by $\mathfrak{T}$ on $X_\lambda$ coincides with $\mathfrak{T}_\lambda$.

In view of Proposition 6 and its corollary, it is enough to show that for each $\lambda$ and each subset $A_\lambda$ of $X_\lambda$ the following statements are equivalent: (i) $A_\lambda$ is open (resp. closed) in the topology $\mathfrak{T}_\lambda$. (ii) For each $\mu \in L$, $A_\lambda \cap X_\mu$ is open (resp. closed) in the topology $\mathfrak{T}_\mu$.

It is clear that (ii) implies (i) by taking $\mu = \lambda$. Conversely, if (i) is satisfied, then $A_\lambda \cap X_\mu$ is open (resp. closed) in $X_\lambda \cap X_\mu$ for the topology $\mathfrak{T}_{\lambda \mu}$ induced on $X_{\lambda} \cap X_{\mu}$ by $\mathfrak{T}_{\lambda}$; but $\mathfrak{T}_{\lambda \mu}$ is also the topology induced on $X_{\lambda} \cap X_{\mu}$ by $\mathfrak{T}_{\mu}$; hence $A_{\lambda} \cap X_{\mu}$ is also the intersection of $X_{\lambda} \cap X_{\mu}$ with a subset $B_{\mu}$ of $X_{\mu}$ which is open (resp. closed) in the topology $\mathfrak{T}_{\mu}$; since $X_{\lambda} \cap X_{\mu}$ is open (resp. closed) in $\mathfrak{T}_{\mu}$, so is $A_{\lambda} \cap X_{\mu}$. This completes the proof.

We remark that if the union of the $X_{\lambda}$ is different from $X$, then the topology induced by $\mathfrak{T}$ on $X - \left( \bigcup_{\lambda \in L} X_{\lambda} \right)$ is discrete. For if $x \in X$ belongs to none of the $X_{\lambda}$, then $\{ x \} \cap X_{\lambda} = \varnothing$ is open in each topology $\mathfrak{T}_{\lambda}$ and therefore $\{ x \}$ is open in the topology $\mathfrak{T}$.