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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?

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    $\begingroup$ An open ball in $\mathbb{R}$ is an open interval. $\endgroup$ – almagest Apr 19 '16 at 17:19
  • $\begingroup$ So what would be the interior of a line segment in $\mathbb{R}^2$? $\endgroup$ – user19405892 Apr 19 '16 at 17:21
  • $\begingroup$ A line segment in $\mathbb R^2$ has empty interion, because an open ball in $\mathbb R^2$ is a circle "without border". $\endgroup$ – gebruiker Apr 19 '16 at 17:23
  • $\begingroup$ But regarded as a subset of $\mathbb{R}$, the interval $[0,1]$ has interior $(0,1)$. $\endgroup$ – almagest Apr 19 '16 at 17:25
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    $\begingroup$ @user19405892: What you mean is called relative interior point. $\endgroup$ – user251257 Apr 19 '16 at 17:35
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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. Three open balls in different metric spaces. One round and two square.

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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.

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