Open interval in $R$ is open set
Let $(a,b) = \{ x \in R \mid a < x < b\}$
Let $c \in (a, b)$ and define $\epsilon$-neighborhood of $c$ as
$N_\epsilon (c) = \{x \in R \mid \ |x - c| < \epsilon \}$
Take $\epsilon < min\{ b - c, c - a\}$.
$x \in N_\epsilon (c) \Longrightarrow |x - c| < \epsilon \Longleftrightarrow c - \epsilon < x < c + \epsilon$
Since $\epsilon < b - c$ and $\epsilon < c - a$,
$ a < x < b $ could be rewritten as
$a = c - (c - a) < c - \epsilon < x < c + \epsilon < b = c - (b - c) \Longleftrightarrow x \in (a, b) \Longrightarrow N_\epsilon (c) \subseteq (a, b)$
Let $c \in (a, b)$ and define $\epsilon$-neighborhood of $c$ as
$N_\epsilon (c) = \{x \in R \mid \ |x - c| < \epsilon \}$
Take $\epsilon < min\{ b - c, c - a\}$.
$x \in N_\epsilon (c) \Longrightarrow |x - c| < \epsilon \Longleftrightarrow c - \epsilon < x < c + \epsilon$
Since $\epsilon < b - c$ and $\epsilon < c - a$,
$ a < x < b $ could be rewritten as
$a = c - (c - a) < c - \epsilon < x < c + \epsilon < b = c - (b - c) \Longleftrightarrow x \in (a, b) \Longrightarrow N_\epsilon (c) \subseteq (a, b)$
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