The interior of the rational numbers topological space is empty

general-topology

I do not quite understand why the interior of the rational numbers topological space is empty, here is my reasoning.

  1. $\mathbb{Q}$ can be regarded as a subspace topology of $\mathbb{R}$ with its usual topology, this means that an open set in $\mathbb{Q}$ will be in the form of $(a,b) \cap \mathbb{Q}$ where $a,b \in \mathbb{R}$ [of course $\mathbb{Q}$ and $\emptyset$ are also open sets in $\mathbb{Q}$].
  2. If $int(\mathbb{Q}) = \emptyset$, then there should not exist an element $x \in \mathbb{Q}$ such that there exists an open subset $U$ of $\mathbb{Q}$ containing $x$.
  3. But, given $x \in \mathbb{Q}$, we can pick any real open interval $(c,d)$ containing $x$, then $(c,d) \cap \mathbb{Q}$ is open in $\mathbb{Q}$ containing $x$ with being a subset of $\mathbb{Q}$, so $x \in int(\mathbb{Q})$ and $int(\mathbb{Q}) \neq \emptyset$.

I am sure some part of my reasoning has to be incorrect, could anyone kindly point it out?

Best Answer

This essentially involves distinguishing subsets from subspaces. The interior of $\mathbb{Q}$ is not empty when $\mathbb{Q}$ is interpreted as the "parent space". Indeed, if $(X,\mathfrak{T})$ is any topological space, then the interior of $X$ (relative to itself) is the entire space $X$!

When we say that $\mathbb{Q}$ has empty interior, we mean that $\mathbb{Q}$ has empty interior as a subset of $\mathbb{R}$. In this case, the basis elements of the topology no longer have the form $(a,b) \cap \mathbb{Q}$. Instead, the basis elements take the form $(a,b)$ which will always contain a point in $\mathbb{R} \setminus \mathbb{Q}$.

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