Let $(X_s, \tau_s)$, for $s\in S$, be topological spaces such that $X_s\neq X_t$ for $s\neq t$; $s,t\in S$, and $X=\bigcup_{s\in S} X_s$.

We define topology $\tau$ in $X$ in the following way: $G\subset X$ is open iff $G\cap X_s \in \tau_s$ for each $s\in S$.

What sequences are convergent in $X$ ?


If you’re assuming that the $X_s$ are pairwise disjoint, the answer is straightforward. Since $\varnothing\in\tau_s$ for every $s\in S$, $X$ is simply the discrete union of the spaces $X_s$: each $X_s$ is a clopen subset of $X$. Thus, a sequence in $X$ converges iff it has a tail that lies entirely in one $X_s$ and converges in that $X_s$.

If you’re not making that assumption, you probably need to say something about how the $\tau_s$ are related on the intersections of the $X_s$.

  • $\begingroup$ Thanks. But is it possible to say something in the case when $S=N$ and $X_s \subset X_{s+1}$ for $s \in N$? Maybe $(x_n)$ is convergent to $x\in X_s$ iff $x_n \in X_s$ for large $n$ and $x_n \rightarrow x$ in $X_s$? $\endgroup$ – Richard Jan 23 '12 at 11:44
  • $\begingroup$ @Richard: Are you assuming in that case that each $\tau_s$ is the relative topology on $X_s$ induced by $\tau_{s+1}$? In that case you could get a sequence converging in $X$ to a point of $X_0$ but with only finitely many terms in any given $X_n$: let $X_0=(\leftarrow,0]$, and for $n>0$ let $X_n=X_0\cup[2^{-n},\to)$, and consider the sequence $\langle 2^{-n}:n\in\mathbb{N}\rangle$. $\endgroup$ – Brian M. Scott Jan 23 '12 at 19:11

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