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
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 | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \set {y: \size {y - 1} < n} \cap \set {y: \size {y + 1} > \dfrac 1 n}\) | \(=\) | \(\ds \openint {1 - n} {-1 - \dfrac 1 n} \cup \openint {-1 + \dfrac 1 n} {1 + n}\) |
$\blacksquare$