# Weak topology and subspaces

Let $$(X_i)_{i \in \mathbb{N}}$$ be a family of topological spaces with inclusion maps $$j_i: X_i \hookrightarrow X_{i+1}$$ (i.e. $$X_i$$ has subspace topology with respect to this maps). We can take their direct limit (in the category of topological spaces) $$X = \varinjlim_{i \in \mathbb{N}} X_i$$ which is the "union" of all $$X_i$$ equipped with the "weak topology", i.e. a set $$A \subset X$$ is open iff its "intersection" with each $$X_i$$ is open in $$X_i$$.

Now we can choose a sequence of subspaces $$Y_i \subset X_i$$ such that $$Y_i \subset Y_{i+1}$$ (or more precisely, $$j_i(Y_i) \subset Y_{i+1}$$). These inclusions give us a morphism between the directed systems $$(X_i)$$ and $$(Y_i)$$, so we get a continuous map between their direct limits $$Y = \varinjlim_{i \in \mathbb{N}} Y_i \to \varinjlim_{i \in \mathbb{N}} X_i = X$$ Clearly this map is injective.

Question: Does $$Y$$ carry the subspace topology with respect to this map?

I don’t have at hand the definition of the limit map between direct the limits, but I guess the question can have a negative answer. Let $$X=\Bbb R^\omega$$ be a subspace of a Tychonoff product $$\Bbb R^\omega$$ consisting of all sequences $$x=(x_i)$$ such that all but finitely many $$x_n$$ are zeroes. Then the space $$X$$ is a direct limit of a sequence of spaces $$X_i$$, where each $$X_i=X$$ and the inclusion maps are the identity maps. For each natural $$i$$ let $$Y_i=\{(x_n)\in\Bbb R^\omega: x_n=0 \mbox{ for all }n>i \}$$ and the inclusion maps are the embeddings. Endow the set $$Y=\bigcup Y_i=\Bbb R^\omega$$ with the topology consisting of all subsets $$U$$ of $$\Bbb R^\omega$$ such that $$U\cap Y_i$$ is open in $$Y_i$$ for each $$i$$. That is, $$Y$$ is a direct limit of the sequence $$\{Y_i\}$$. Now let $$a^n=(a^n_i)$$ be a sequence of elements of $$\Bbb R^\omega$$ such that for each $$n$$ and $$i$$ we have $$a^n_i$$ equals $$1$$, if $$n=i$$, and equals $$0$$, otherwise. Then the sequence $$(a^n)$$ converges to $$0$$ in $$X$$, but not in $$Y$$.