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Let $f$ be a function from a metric space $X$ to a metric space $Y$. Show that if $f$ is uniformly continuous on $X$ then $f$ is continuous on $X$. Show that the converse is not true.

Uniform continuity definition:

$∀ε>0:∃δ>0:∀p,q∈X:d_{X}(p,q)<δ⟹d_{Y}(f(p),f(q))<ε$

Continuity definition:

$∀ε>0:∀p∈X:∃δ>0:∀q∈X:d_{X}(p,q)<δ⟹d_{Y}(f(p),f(q))<ε$

I wonder if these proofs are correct and if they are formal enough?


Uniform continuity ⟹ Continuity

  1. Let $f$ be uniformly continuous. Fix $ε_{0}$, obtain $δ_{0}(ε_{0})$ (as a function of $ε_{0}$), fix any $p_{0}$ and $q_{0}$ and we know that:

$d_{X}(p_{0},q_{0})<δ_{0}⟹d_{Y}(f(p_{0}),f(q_{0}))<ε_{0}$

  1. If we want to prove that $f$ is continuous at $p_{0}$, we fix $ε_{0}$ and we pick the same $δ_{0}$ as above and fix any $q_{0}$ and we are assured by 1. that

$d_{X}(p_{0},q_{0})<δ_{0}⟹d_{Y}(f(p_{0}),f(q_{0}))<ε_{0}$

  1. Since $p_{0}$ was arbitrary, $f$ is continuous at $X$.

Continuity $\color{red}{\implies}$ Uniform continuity

  1. Let $f$ be continuous. Fix $ε_{0}$, fix any obtain $p_{0}$ and $p_{1}$, obtain both $δ_{0}(ε_{0},p_{0})$ and $δ_{1}(ε_{0},p_{1})$ (not necessarily equal), fix any $q_{0}$ and $q_{1}$ and we know that:

$d_{X}(p_{0},q_{0})<δ_{0}⟹d_{Y}(f(p_{0}),f(q_{0}))<ε_{0}$

$d_{X}(p_{1},q_{1})<δ_{1}⟹d_{Y}(f(p_{1}),f(q_{1}))<ε_{0}$

  1. For $f$ to be uniformly continuous, if we fix $ε_{0}$, obtain $δ$, fix any $p_{0}$, $p_{1}$, $q_{0}$ and $q_{0}$ we must have that:

$d_{X}(p_{0},q_{0})<δ⟹d_{Y}(f(p_{0}),f(q_{0}))<ε_{0}$

$d_{X}(p_{1},q_{1})<δ⟹d_{Y}(f(p_{1}),f(q_{1}))<ε_{0}$

  1. But those two implications need not hold for the same $δ$ as shown above. Suppose we pick $δ_{0}$ and w.l.o.g. we suppose $δ_{0}>δ_{1}$. Then it could be the case that

$δ_{1}<d_{X}(p_{1},q_{1})<δ_{0}$

but we cannot conclude ⟹$d_{Y}(f(p_{1}),f(q_{1}))<ε_{0}$.

Therefore, continuity does not imply uniform continuity.

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  • $\begingroup$ I think your wording could be a little clearer on the first one $\endgroup$ – KyleW Jul 29 '15 at 23:04
  • $\begingroup$ @KyleW I agree. As a side note, I discovered that $\exists y: \forall x: \phi(x,y)$ implies $\forall x: \exists y: \phi(x,y)$ but not the other way round. If we fix $ε$ and $q$, the first and last universal quantifier respectively, we are left with $\exists δ: \forall p\in X: \phi(δ,p)$, then $ \forall p\in X: \exists δ: \phi(δ,p)$. $\endgroup$ – julian.marr Jul 29 '15 at 23:30
  • $\begingroup$ When you do the proof pedantically you find that that is basically all that you are proving is that implication $\endgroup$ – KyleW Jul 29 '15 at 23:41
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You cannot usually prove than an implication fails; what you do is provide a counterexample. Your argument for continuity does not imply uniform continuity cannot work because (as you showed before) there are functions which are both.

For a continuous but not uniformly continuous function, intuitively you need a function that has arbitrarily high slopes (i.e., unbounded derivative if it has one). An easy example is $f(x)=x^2$ with $X=Y=\mathbb R$. We need to show that this fails the definition of uniform continuity: that is, it satisfies its negation, which says $$\tag{1} \exists\varepsilon>0, \forall\delta>0,\ \exists p,q:\ d(p,q)<\delta\ \text{ and }d(f(p),f(q))>\varepsilon. $$ For this function in particular we can fix any $\varepsilon$, say $\varepsilon=1$. Given $\delta>0$, we can take $p=k$, $q=k+\delta/2$ for some integer $k$ (to be determined). Then $d(p,q)=|p-q|=\delta/2$, and $$ d(f(p),f(q))=d(p^2,q^2)=|(k+\delta/2)^2-k^2|=2k\delta+\delta^2/4>2k\delta. $$ So if we choose $k>1/2\delta$, $(1)$ is satisfied.

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