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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