1. ## Uniform Continuity definition

Let $(M,d)$ and $(N,p)$ be metric spaces, and let $f:A \subset M \rightarrow N$. Prove that f is an uniformly continuous mapping on A iff for every pair of sequence $\{ x_k \} , \{ y_k \}$ of A with $d(x_k,y_k) \rightarrow 0$, we have $p(f(x_k),f(y_k)) \rightarrow 0$

Proof.

Suppose that f is uniformly continuous, pick sequences $\{ x_k \} , \{ y_k \}$ of A such that $d( x_k,y_k) \rightarrow 0$

Let $\epsilon > 0$ be given, and let $\delta > 0$, then $\exists K \in \mathbb {N}$ such that whenever $k \geq K$, we have $d(x_k,y_k) < \delta$. Since f is uniformly continuous, we then ahve $p(f(x_k),f(y_k)) < \epsilon$, implies that $p(f(x_k),f(y_k)) \rightarrow 0$

Conversely, suppose that math] \{ x_k \} , \{ y_k \} [/tex] are sequences in A with $d( x_k,y_k) \rightarrow$, we have $p(f(x_k),f(y_k)) \rightarrow 0$.

Let $\epsilon > 0$ be given, pick $\delta > 0$, find $K \in \mathbb {N}$ such that whenever $k \geq K$, we have $d(x_k,y_k) < \delta$.

Now suppose that $x_k \rightarrow x_0$ and $y_k \rightarrow y_0$, then $d(x_0,y_0) < \delta$, but we also have $p(f(x_k),f(y_k)) \rightarrow 0$, which means $p(f(x_k),f(y_k)) < \epsilon$.

Q.E.D.

Is this right? Thanks!

Let $(M,d)$ and $(N,p)$ be metric spaces, and let $f:A \subset M \rightarrow N$. Prove that f is an uniformly continuous mapping on A iff for every pair of sequence $\{ x_k \} , \{ y_k \}$ of A with $d(x_k,y_k) \rightarrow 0$
Clearly if $f$ is uniformly continuous the statement is correct but not your proof.
From U.C. $\varepsilon > 0 \Rightarrow \left( {\exists \delta > 0} \right)\left[ {d(x,y) < \delta \Rightarrow p\left( {f(x),f(y)} \right) < \varepsilon } \right]\quad$.
Suppose that $\left( {x_n } \right) \to x\,\& \,\left( {y_n } \right) \to y$ then $
\left( {\exists N} \right)\left[ {n \geqslant N \Rightarrow d\left( {x_n ,x} \right) < \frac{\delta }{3} \wedge d\left( {y_n ,y} \right) < \frac{\delta }{3}} \right]$
.
Now consider this: If $d(x,y) < \frac{\delta }{3}$ then $d\left( {x_n ,y_n } \right) \leqslant d\left( {x_n ,x} \right) + d\left( {x,y} \right) + d\left( {y,y_n } \right) < \delta \Rightarrow p\left( {f\left( {x_n } \right),f\left( {y_n } \right)} \right) < \varepsilon
$
.

P.S. I was far too tired to think straight last night.

The converse is easy. If $f$ is not informally continuous then $\left( {\exists \lambda > 0} \right)\left( {\forall \delta > 0} \right)\left( {\exists x_\delta \wedge y_\delta } \right)\left[ {d(x_\delta ,y_\delta ) < \delta \wedge p(f(x_\delta ),f(y_\delta )) \geqslant \lambda } \right]$.
That is the negation of informally continuous.

So $\left( {\forall n} \right)\left( {\exists x_n \wedge y_n } \right)\left[ {d(x_n ,y_n ) < \frac{1}{n} \wedge p(f(x_n ),f(y_n )) \geqslant \lambda } \right]$
That is a contradiction, because $d(x_n ,y_n ) \to 0$.