Size of y-1 lt n and Size of y+1 gt 1 over n

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Theorem

Let $T_n \subseteq \R$ be the subset of the set of real numbers $\R$ defined as:

$T_n = \set {y: \size {y - 1} < n \land \size {y + 1} > \dfrac 1 n}$

Then:

$T_n = \openint {1 - n} {-1 - \dfrac 1 n} \cup \openint {-1 + \dfrac 1 n} {1 + n}$


Proof

Union-of-Family-Example-1.png

First note that:

\(\ds T_n\) \(=\) \(\ds \set {y: \size {y - 1} < n \land \size {y + 1} > \dfrac 1 n}\)
\(\ds \) \(=\) \(\ds \set {y: \size {y - 1} < n} \cap \set {y: \size {y + 1} > \dfrac 1 n}\)


We have:

\(\ds \set {y: \size {y - 1} < n}\) \(=\) \(\ds \openint {1 - n} {1 + n}\) Open Interval Defined by Absolute Value
\(\ds \set {y: \size {y + 1} > \dfrac 1 n}\) \(=\) \(\ds \R \setminus \closedint {-1 - \dfrac 1 n} {-1 + \dfrac 1 n}\) Complement of Closed Interval Defined by Absolute Value
\(\ds \leadsto \ \ \) \(\ds \set {y: \size {y - 1} < n} \cap \set {y: \size {y + 1} > \dfrac 1 n}\) \(=\) \(\ds \openint {1 - n} {-1 - \dfrac 1 n} \cup \openint {-1 + \dfrac 1 n} {1 + n}\)

$\blacksquare$