Prove that $\mathbb{Q}$ with the topology induced by the usual distance is not complete. My question is not really about proving this fact. Instead, it is about the sense of the sentence. What does "the topology induced by the usual distance" mean? I never understood it clearly.
To prove the proposition, I thought of the Cauchy sequence defined by $a_n=\left(1+\frac{1}{n}\right)^{n}$, which converges to $e$ and does not belong to $\mathbb{Q}$.
Thanks!
 A: So first let us start by defining the usual distance on the real numbers.

Definition 1:  The Standard Distance on $\mathbb{R}$, also called the usual distance is defined by $\forall a\in \mathbb{R}, \forall b\in \mathbb{R}$
$$d(a,b)=\vert a-b \vert$$

That same metric can be considered over $\mathbb{Q}$.
Now that we have a metric on $\mathbb{Q}$ we can create a topology by defining a basis. We start by taking for any point $a\in \mathbb{Q}$ the open interval centered around $a$ to be topologically open i.e. for any positive real number $\epsilon$
$$(a-\epsilon,a+\epsilon)$$
Now we define open sets. We say that a set $A$ is open iff it can be created by taking the union of collection of open intervals. This may be a collection of finite open sets or infinitely many open intervals.

Note We use the term usual metric because this metric reproduces all the normal results we use in calculus suich as open intervals, closed intervals, limits of a sequence, completeness etc.. Which means using the standard metric, you can use what you know about those properties from calculus without reproving them.

Now for completeness

Definition 2: Let $S=\{x_1, x_2, x_3,\dots\}$ be an infinite sequence in a $(\mathbb{Q},d)$. We say that $S$ is complete iff  it has a limit in $\mathbb{Q}$ i.e. $$\lim_{i\to +\infty} x_i=x\in \mathbb{Q}$$

Now to prove that $(\mathbb{Q},d)$ is not complete.
Consider the sequence $S=\{3, 3.1, 3.14, 3.141, 3.1415, \dots\}$ basically the sequence where $x_n$ has the first $n$ decimal places of $\pi$. Clearly
$$\lim_{i\to +\infty} x_i=\pi$$
but $\pi\notin \mathbb{Q}$ which means that technically, $S$ has not limit in $\mathbb{Q}$.
Since we found a sequence in $(\mathbb{Q},d)$ that has no limit in $(\mathbb{Q},d)$, we deduce that $(\mathbb{Q},d)$ is not complete.
Hope this helped.
