We consider a direct limit of a tower $X_1\subset\cdots\subset X_n$ of spaces, where each $X_n$ is a subspace of $X_{n+1}$. The direct limit is $X_{\infty}:=\cup_n X_n$ endowed with the topology $\mathcal{T}_{\infty}$ defined as follows: $U\subset X_{\infty}$ is open if and only if $\forall n\in \Bbb{N},\quad U\cap X_n$ is open on $\mathcal{T_n}.$

We assume that $\mathcal{T}_n$ is the subspace topology on $X_n$ by $\mathcal{T_{n+1}}.$

I would like to prove that the subspace topology on $X_n$ by $\mathcal{T}_{\infty}$ denoted $\mathcal{T'}$ is $\mathcal{T}_n.$

Clearly we have $\mathcal{T'}\subset \mathcal{T_n}.$

For the converse, let $\omega_0\in \mathcal{T_n}$ which can be written as $\omega_0=\omega_1\cap X_n$ with $\omega_1\in \mathcal{T}_{n+1}.$

Using the fact that $X_n\subset X_{n+1}$ I can write $\omega_0=\omega_1\cap X_n=\omega_2\cap X_{n}$ with $\omega_2\in \mathcal{T}_{n+2}$ and so on. But I would like to prove that $\omega_0=\omega_{\infty}\cap X_n$ where $\omega_{\infty}\in \mathcal{T}_{\infty}$

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Let's do it for $n=1$

Let $\omega_1\in\mathcal T_1$.

Then $\omega_1=\omega_2\cap T_1$ for some $\omega_2\in\mathcal T_2$.

Then $\omega_2=\omega_3\cap T_2$ for some $\omega_3\in\mathcal T_3$.

Et cetera.

We have: $\omega_1\subseteq\omega_2\subseteq\omega_3\subseteq\cdots$

For a fixed $n$ it can be shown by induction that $\omega_{n+k}\cap X_n=\omega_n$ for $k=0,1,2,\dots$

Then for $\omega:=\bigcup_{n=1}^{\infty}\omega_n$ we find $\omega\cap X_n=\omega_n\in \mathcal T_n$ for every $n$ so we conclude that $\omega\in\mathcal T_{\infty}$.

Consequently $\omega_1=\omega\cap X_1\in\mathcal T'$.

This procedure will also work for $n=2,3,\dots$

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