# Does $\mathcal B([0,1])=\{U\cap [0,1]\mid U\in \mathcal B(\mathbb R)\}$?

Endow the subset $$X$$ of $$\mathbb{R}$$ with the subspace topology. Denote $$\mathcal B(X)$$ as the set of all the Borel sets of $$X$$. Does one have $$\mathcal B([0,1])\underset{(1)}{=}\{U\cap [0,1]\mid U\in \mathcal B(\mathbb R)\}\ ?$$

For (1) I tried as follow. We have that $$\{U\cap [0,1]\mid U\in \mathcal B(\mathbb R)\},$$ is a $$\sigma -$$algebra (easy). Since open set of $$[0,1]$$ are of the form $$[0,1]\cap O$$ where $$O$$ is open in $$\mathbb R$$, we have that $$\mathcal B([0,1])\subset \{U\cap [0,1]\mid U\in \mathcal B(\mathbb R)\}.$$ How can I prove the converse inclusion ?

There is a general result:

$$\textbf{Theorem}$$ Let $$\mathcal{F}$$ be a collection of subsets of $$X$$ and $$Y \subseteq X$$ non-empty.Denote $$\sigma(\mathcal{F})_Y=\{A \cap Y:A \in \sigma(\mathcal{F})\}$$ $$\mathcal{F}_Y=\{A \cap Y:A \in \mathcal{F}\}$$Then $$\sigma(\mathcal{F})_Y=\sigma(\mathcal{F}_Y)$$ where $$\sigma(B)$$ is the sigma algebra generated by the collection of sets $$B$$

$$\textbf{Proof}$$

Let $$S \in \mathcal{F}_Y$$ then $$S=A \cap Y$$ for some $$A \in \mathcal{F} \subseteq \sigma(\mathcal{F})\Longrightarrow S \in \sigma(\mathcal{F})_Y$$ and since $$\sigma(\mathcal{F})_Y$$ is a sigma algebra(easy exercise for you) we have that $$\sigma(\mathcal{F}_Y) \subseteq \sigma(\mathcal{F})_Y$$

Now define the collection $$\Sigma=\{A \subseteq X:A \cap Y \in \sigma(\mathcal{F}_Y)\}$$

We have that $$\Sigma$$ is a sigma algebra of subsets of $$X$$(again an easy exercise for you).

If $$A \in \mathcal{F}$$ then $$A \cap Y\in \mathcal{F}_Y \subseteq\sigma(\mathcal{F}_Y)$$

hence $$A \in \Sigma$$

Therefore $$\mathcal{F} \subseteq \Sigma\Longrightarrow \sigma(\mathcal{F}) \subseteq \Sigma$$

Now for an arbitrary $$B \in \sigma(\mathcal{F})_Y$$ we have that $$B=A \cap Y$$ for some $$A \in \sigma(\mathcal{F}) \subseteq \Sigma$$ and thus $$B \in \sigma(\mathcal{F}_Y)$$

This implies that $$\sigma(\mathcal{F})_Y \subseteq \sigma(\mathcal{F}_Y)$$

Now apply this theorem with $$\mathcal{F}=$$open sets(topology of $$\Bbb{R}$$) and $$Y=[0,1]$$ and $$X=\Bbb{R}$$

Note that $$B[0,1]=\{A \subseteq [0,1]:A \in B(\Bbb{R})\}$$ and $$[0,1]$$ has the subspace topology.