# Show that if $g((x_n)) \rightarrow l$ and $g((y_n)) \rightarrow m$, then $l=m$

Suppose that $g: (a,b] \rightarrow \mathbb{R}$ is uniformly continuous. Suppose that both $(x_n), (y_n)$ are sequence in $(a,b]$ which converge to $a$. Show that if $g(x_n)) \rightarrow l$ and $g((y_n)) \rightarrow m$, then $l=m$.

By definition, since $(x_n) \rightarrow a$, $\forall \varepsilon > 0$, $\exists N_0 \in \mathbb{N}$ $\forall n\geq N_0$

$|x_n-a|<\varepsilon$

A parallel statement can be made for the sequence $(y_n)$. This implies that

$|x_n-y_n|=|x_n-a+a-y_n| \leq |x_n-a|+|a-y_n| \equiv |x_n-a|+|y_n-a| < \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$

Since $g$ is uniformly continuous, this informs us that $\forall \varepsilon>0$, $\exists \delta>0$ s.t. $\forall x_n \in (x_n)$ and $\forall y_n \in (y_n)$,

$|g(x_n)-g(y_n)|<\varepsilon$ (we exclude the delta condition since we showed that it is always satisfied)

Similar conditions can be made using each sequence exclusively

$|g(x_n)-g(x_m)|<\varepsilon$

$|g(y_n)-g(y_m)|<\varepsilon$

Struggling to figure out the last few steps.

• Your proof needs work. Too many "$\forall \varepsilon > 0$, $\exists N_0 \in \mathbb{N}$ $\forall n\geq N_0$" statements. Suggestion: Start with "let $\epsilon > 0.$" Now, it seems you have the basic idea for proving $x_n-y_n \to 0.$ But this also converges to ___ .
– zhw.
Apr 20, 2015 at 19:12

Here is a proof in the framework of IST, assuming $g$, $a$, $l$ and $m$ are standard. Let $n$ be infinitely large. Then $x_n$ and $y_n$ are both infinitely close to $a$, so $x_n$ is infinitely close to $y_n$. By uniform continuity, $g(x_n)$ is infinitely close to $g(y_n)$. Because $g(x_n)\to l$, $g(x_n)$ is infinitely close to $l$ and similarly $g(y_n)$ is infinitely close to $m$. Thus $l$ and $m$ are infinitely close, hence (being standard) equal.