In order to elaborate on Joel Hamkins comment: The $\sigma$ algebra $A$ generated by the open sets of $\mathbb{R}$ is not the power set of $\mathbb{R}$, since there are subsets of $\mathbb{R}$, which are not contained in $A$ by the axiom of choice. Now suppose $A$ is a topology, then for every subset $X \subset \mathbb{R}$ as the union $\cup_{x \in X} \\left{ x \right}$ would be in the topology, and hence measurable, which contradicts the observation that $ A \neq P(\mathbb{R})$.

In order to elaborate on Joel Hamkins comment: The $\sigma$ algebra $A$ generated by the open sets of $\mathbb{R}$ is not the power set of $\mathbb{R}$, since there are subsets of $\mathbb{R}$, which are not contained in $A$ by the axiom of choice. Now suppose $A$ is a topology, then for every subset $X \subset \mathbb{R}$ as the union $\cup_{x \in X} \left\{ x \right\}$ would be in the topology, and hence measurable, which contradicts the observation that $ A \neq P(\mathbb{R})$.

In order to elaborate on Joel Hamkins comment: The $\sigma$ algebra $A$ generated by the open sets of $\mathbb{R}$ is not the power set of $\mathbb{R}$, since there are subsets of $\mathbb{R}$, which are not contained in $A$ by the axiom of choice. This equivalent to existence of non Lebesgue measurable sets. Now suppose $A$ is a topology, then for every subset $X \subset \mathbb{R}$ as the union $\cup_{x \in X} \\left{ x \right}$ would be in the topology, and hence measurable, which contradicts the observation that $ A \neq P(\mathbb{R})$.

In order to elaborate on Joel Hamkins comment: The $\sigma$ algebra $A$ generated by the open sets of $\mathbb{R}$ is not the power set of $\mathbb{R}$, since there are subsets of $\mathbb{R}$, which are not contained in $A$ by the axiom of choice. Now suppose $A$ is a topology, then for every subset $X \subset \mathbb{R}$ as the union $\cup_{x \in X} \\left{ x \right}$ would be in the topology, and hence measurable, which contradicts the observation that $ A \neq P(\mathbb{R})$.