# Borel $\sigma$-algebras and topologies

If $id:(M,\rho_1)\to(M,\rho_2)$ is continuous, then $\cal{B}_2$$(M)\subseteq \cal{B}_1$$(M)$

where $\cal{B}_1, \cal{B}_2$ are the Borel $\sigma$-algebras induced by the metrics $\rho_1 ,\rho_2$ respectively.

I've understood that by continuity of f, since the pre-image of an open set is open, then open sets $U$ that are open w.r.t $\rho_2$ are also open w.r.t $\rho_1$. This means that if $\tau_1,\tau_2$ are topologies w.r.t $\rho_1, \rho_2$ then $\tau_2 \subseteq \tau_1$

But how am I able to obtain/prove that $\cal{B}_2$$(M)\subseteq \cal{B}_1$$(M)$ must follow?

• If $U$ is open w.r.t. $\rho_2$, then $f^{-1}(U)$ is open w.r.t. $\rho_1$. How does this imply $U$ is open w.r.t. $\rho_1$? – parsiad May 5 '17 at 17:48
• Regardless, if you know $\tau_2 \subset \tau_1$, then trivially $\mathcal{B}_2(M) \subset \mathcal{B}_1(M)$ since $\mathcal{B}_i(M)=\sigma(\tau_i)$ where $\sigma(X)$ is the sigma algebra generated by $X$. – parsiad May 5 '17 at 17:49
• I forgot to add that f is the identity function in this case – Malcolm May 5 '17 at 17:55
• Is there someway to prove that if $\tau_2 \subseteq \tau_1 \Rightarrow \sigma(\tau_2) \subseteq \sigma(\tau_1)$? I'm not at the stage where I can visualize it – Malcolm May 5 '17 at 17:57
• Clearly $\tau_2 \subseteq \sigma(\tau_1)$. Hence $\sigma(\tau_1)$ is a $\sigma$-algebra which contains $\tau_2$. By definition, $\sigma(\tau_2)$ is the smallest $\sigma$-algebra containing $\tau_2$, so necessarily it is smaller than $\sigma(\tau_1)$, i.e. we have $\sigma(\tau_2) \subseteq \sigma(\tau_1)$. – Nate Eldredge May 5 '17 at 18:02

Theorem: If $X \subset Y$, then $\sigma(X) \subset \sigma(Y)$.
Proof: Since $$\sigma(X)\equiv\bigcap_{\substack{\mathcal{M}\text{ is a }\sigma\text{-algebra}\\ X\subset\mathcal{M} } }\mathcal{M} \text{ and } \sigma(Y)\equiv\bigcap_{\substack{\mathcal{M}\text{ is a }\sigma\text{-algebra}\\ Y\subset\mathcal{M} } }\mathcal{M},$$ the desired result follows.