Inequality Rule for Real Convergent Nets

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Theorem

Let $\struct {\Lambda, \preceq}$ be a directed set.

Let $\family {x_\lambda}_{\lambda \mathop \in \Lambda}$ and $\family {y_\lambda}_{\lambda \mathop \in \Lambda}$ be indexed families of elements in $\R$.

Let $\family {x_\lambda}_{\lambda \mathop \in \Lambda}$ and $\family {y_\lambda}_{\lambda \mathop \in \Lambda}$ be convergent to the following limits:

$\ds \lim_{\lambda \mathop \in \Lambda} x_\lambda = l$

$\ds \lim_{\lambda \mathop \in \Lambda} y_\lambda = m$

Let there exist $\lambda \in \Lambda$ such that:

$\forall \mu \in \Lambda : \lambda \preceq \mu: x_\mu \le y_\mu$

Then:

$l \le m$

Proof

Aiming for a contradiction, suppose

$l > m$

Let $\epsilon = \dfrac {\paren{l - m}} 2$.

Hence:

$\epsilon > 0$

and:

$(1) \quad l - \epsilon = m + \epsilon$

By definition of convergence:

$\exists \lambda_1 \in \Lambda : \forall \mu \in \Lambda : \lambda_1 \preceq \mu \implies \size{l - x_\mu} < \epsilon$

and:

$\exists \lambda_2 \in \Lambda : \forall \mu \in \Lambda : \lambda_2 \preceq \mu \implies \size{m - y_\mu} < \epsilon$

From Closed Interval Defined by Absolute Value:

$(2) \quad \forall \mu \in \Lambda : \lambda_1 \preceq \mu \implies l - \epsilon < x_\mu$

and:

$(3) \quad \forall \mu \in \Lambda : \lambda_2 \preceq \mu \implies m + \epsilon > y_\mu$

By definition of directed set:

$\exists \mu \in \Lambda : \lambda, \lambda_1, \lambda_2 \preceq \mu$

We have:

\(\ds x_\mu\)

\(>\)

\(\ds l - \epsilon\)

from $(2)$

\(\ds \)

\(=\)

\(\ds m + \epsilon\)

from $(1)$

\(\ds \)

\(>\)

\(\ds y_\mu\)

from $(3)$

This contradicts the hypothesis:

$\forall \mu \in \Lambda : \lambda \preceq \mu: x_\mu \le y_\mu$

Hence:

$l \le m$

$\blacksquare$