# Increasing sequence of topological spaces

Let $\{(X_{n},\tau_{n})\}_{n\in \mathbb{N}}$ be a sequence of topological spaces, such that for every $n \in \mathbb{N}$, $(X_{n}, \tau_{n})$ is a subspace of $(X_{n+1}, \tau_{n+1})$ for all $n \in \mathbb{N}$. Let $X = \bigcup_{n \in \mathbb{N}} X_{n}$.

Let $\tau = \lbrace A \subseteq X | (\forall n \in \mathbb{N}) A \cap X_{n} \in \tau_{n} \rbrace$.

1. Prove that $(X, \tau)$ is a topological space.

2. Prove that $(X_{n}, \tau_{n})$ is a subspace of $(X, \tau)$ for every $n \in \mathbb{N}$.

3. Prove that a function $f:X \to Y$ is continuous ($Y$ is some topological space) iff $f|_{X_n}$ is continuous for all $n \in \mathbb{N}$.

4. Prove that $(X, \tau)$ is $T_{1}$ iff $(\forall n \in \mathbb{N}) (X_{n}, \tau_{n})$ is $T_{1}$.

I've proven 1. and 3. pretty easily, the first by definition, and the third by noticing that for any $B \in \tau_{Y}$, $f^{-1}(B) = \bigcup_{n \in \mathbb{N}} f^{-1}|_{X_{n}}(B)$.

However, I'm having trouble with 2. and 4. For 2, I need to prove the following equivalence: $A \in \tau_{n} \iff \exists B \in \tau : A = B \cap X_{n}$. I'm having trouble with $\implies$: I've tried picking $B_{m} \in \tau_{m}$ for $m \geq n$ such that $A=B_{m} \cap X_{n}$, and then setting $B = \bigcup_{m \geq n} B_{m}$, but I can't prove that $B \cap X_{j} \in \tau_{j}$ for all $j$ (if $j>n$, then I can't prove that $B \cap X_{j} = \bigcup_{m \in \mathbb{N}} (B_{m} \cap X_{j})$ is open in $\tau_{j}$, because for $m < j$, I don't know if $B_{m}$ are open in $\tau_{j}$).

As far as 4, I haven't gotten any further than applying the definition directly: if $x, y \in X$, $x \neq y$, then $x, y \in X_{m}$ starting with $m= m_{0}$. So then if $X_{m}$ are $T_{1}$, I get $U_{m}$, $V_{m}$ such that $x \in U_{m} \setminus V_{m}$, $y \in V_{m} \setminus U_{m}$. So I'd like to have $\bigcup_{m \geq m_{0}} U_{m}$ and $\bigcup_{m \geq m_{0}} U_{m}$ be open in $\tau$, but I don't know how to prove that either.

• For 3) suppose each $X_n$ is $T_1$ and let $x \neq y$ in X. There exists an integer N such that x and y both belong to $X_N$. ( The fact that $X_n \subset X_{n+1}$ is used here). Let U be an open set in $X_N$ containing x but not y. Then U is also open in X and it contains x but not y. Similarly there exists an open set V in X containing y but not x. Converse part: suppose X is $T_1$. Fix N and let $x \neq y$ in $X_N$. Let U be an open set in X containing x but not y. Then $U \cap X_N$ is open in $X_N$, it contains x but not y. Similarly there is an open set in $X_N$ containing y but not x. – Kavi Rama Murthy Jan 10 '18 at 5:53

For part 2, choose the $B_i$ nested, where $B_n=B_{n+1}\cap X_n$. This choice is possible since each $X_n$ is a subspace of $X_{n+1}$. It then follows that $B\cap X_j=B_j$.
For part 4, recall that $T_1$ is equivalent to points being closed, since clearly if points are closed then $V=X\setminus \{x\}$ and $U=X\setminus \{y\}$ are the desired open sets, and for $X$ $T_1$ we have for each $y\in X$, $y\neq x$, an open $U (y)$ not containing $x$, and hence $\bigcup_{y\neq x}U (y)=X\setminus \{x\}$ is open.
Now, let $U=X\setminus \{x\}$. Then $U\cap X_n= X_n\setminus \{x\}$ which is open by assumption, so $U$ is open and therefore $\{x\}$ is closed.
• If $X$ is $T_1$ all its subspaces are, so all $X_n$ are. Just as a remark for the other direction of 4. – Henno Brandsma Jan 10 '18 at 5:27