Let $(X_n)_{n \in \mathbb{N}}$ be a collection of topological spaces, let $X$ be a set and let $X_n \hookrightarrow X$ be a collection of injective maps. Consider in $X$ the final topology given by these maps, so

$G \subseteq X$ is open $\iff$ $G \cap X_n \subseteq X_n$ is open for all $n$.

Now consider $Y \subseteq X$ be a closed subspace, and set $Y_n := Y \cap X_n$.

Question. Is it true that the subspace topology on $Y$ is the same that the final topology given by $Y_n \hookrightarrow Y$? In other words,

$G \subseteq Y$ is open $\iff$ $G \cap Y_n \subseteq Y_n$ is open for all $n$.

  • $\begingroup$ No, you need to relate the Y_n injections to X_n injections. The notation for those injections is terrible. Let f_n:X_n -> X, n in N be a collection of injections. Using f_ n is easier to work with. Use it. $\endgroup$ – William Elliot Nov 7 '17 at 21:13

Obviously, both statement for open sets are true iff they are true for closed sets. Ie, the problem can be translated to

$$U \subseteq Y \text{ closed} \iff U \cap Y_n \text{ closed } \forall n \in \mathbb{N}$$

$(\Rightarrow ).$ It is the definition of subspace topology, nothing exciting.

$(\Leftarrow ).$ Since $Y$ is closed in $X$, then $Y_n = Y \cap X_n$ is closed in $X_n$ for all $n\in \mathbb{N}$. But as the $U \cap Y_n$ is closed in $Y_n$, and we said that $Y_n$ is closed in $X_n$, then (recall that a subset that is closed in a closed subspace of a topological space is closed in the whole space ) $U \cap Y_n = ( U \cap Y ) \cap X_n $ is closed in $X_n$ for all $n$. Having $X$ the final topology given by $X_n \hookrightarrow X$, this is equivalent to $U \cap Y$ being closed in $X$, thus $U= U \cap Y = U \cap Y \cap Y$ is closed in $Y$. $\blacksquare$


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