Definition of interior

Question

An interior point is defined as the following in the Euclidean space.

If $S$ is a subset of a Euclidean space, then $x$ is an interior point of $S$ if there exists an open ball centered at $x$ which is completely contained in $S$.

But doesn't this definition contradict the following?

If $X$ is the Euclidean space $\mathbb{R}$ of real numbers, then $\text{int}([0, 1]) = (0, 1).$

Shouldn't the interior of that be empty since there are no open balls centered at a point $x$ on $(0,1)$ which are completely contained in the line segment?

Answer 1

If we have a metric space $(M,d)$, then an open ball with centre $x$ and radius $\varepsilon$ is the set $$B_\varepsilon(x):=\{y\in M\mid d(x,y)<\epsilon\}.\tag{1}$$ Each time you are dealing some particular metric space $(M,d)$, you should start over and see what $B_\varepsilon(x)$ actually represents, by just writing out the definion.

In the OP the metric space is $(\mathbb R,\vert\cdot\vert)$. The line segment $[0,1]$ is now a subspace of $\mathbb R$.

In the comments the OP asks about a line segment in $\mathbb R^2$. Here we live in the metric space $(\mathbb R^2,\Vert\cdot\Vert)$. Though in both cases we consider line segments, they are considered to be subsets of different spaces.

N.B: Don't get confused by the word ball. The open ball $B_\epsilon(x)$ is just the set defined in $(1)$. It is very common that it is not actually something round. See for instance the picture below. These are all open balls in different metric spaces.

Answer 2

In $\mathbb{R}$, an open ball of radius $r>0$ about a point $x$ is a set of the form $B(x;r):=\{y\in \mathbb{R}| |y-x|<r\}=(x-r,x+r)$. In other words, in $\mathbb{R}$, open balls are just open intervals.