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Let $X$ be a subset of $\mathbb{R}$, and let $f : X\to \mathbb{R}$ be a function. Then the following two statements are logically equivalent:

(a) $f$ is uniformly continuous on $X$.

(b) Whenever $(x_n)$ and $(y_n)$ are two equivalent sequences consisting of elements of $X$, the sequences $(f(x_n))$ and $(f(y_n))$ are also equivalent.

Proof

First I will state the definitions of uniform continuity and equivalent sequences.

(Uniform continuity). Let $X$ be a subset of $\mathbb{R}$, and let $f : X\to\mathbb{R}$ be a function. We say that $f$ is uniformly continuous if, for every $\epsilon > 0$, there exists a $\delta > 0$ such that $f(x)$ and $f(x_0)$ are $\epsilon$-close whenever $x, x_0 \in X$ are to points in $X$ which are $\delta$-close.

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(Equivalent sequences). Let $m$ be an integer, let $( a_n)_{n=m}^\infty$ and $( b_n)_{n=m}^\infty$ be two sequences of real numbers, and let $\epsilon> 0$ be given. We say that $( a_n)_{n=m}^\infty$ is $\epsilon$-close to $( b_n)_{n=m}^\infty$ iff $a_n$ is $\epsilon$-close to $b_n$ for each $n\geq m$. We say that $( a_n)$ is eventually $\epsilon$-close to $( b_n)$ iff there exists an $N\geq m$ such that the sequences $(a_n)$ and $(b_n)$ are $\epsilon$-close. Two sequences $(a_n)$ and $(b_n)$ are equivalent iff for each $\epsilon> 0$, the sequences $(a_n)$ and $(b_n)$ are eventually $\epsilon$-close.

since $x\in X$ is an adherent point to $X$, then there exists a sequence $(a_n)$ such that $a_n \in X$ and converges to x. since f is continuous, then the sequence $(f(a_n))$ converges to $f(x)$.

Let $(b_n)$ be a sequence equivalent to $(a_n)$. Therefore, $\forall \epsilon>0, \exists N\text{ such that } |a_n-b_n|\leq\epsilon$. choose $\epsilon=\delta$, then we have $|a_n-b_n|\leq\delta \forall n\geq N$

Hence, $|f(a_n)-f(b_n)|\leq\epsilon\text{ }, \forall n\geq N$

hence, $(f(a_n))$ and $(f(b_n))$ are equivalent.

Is my proof correct?

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    $\begingroup$ It seems that your proof is nearly correct that $a \Rightarrow b$, though I don't know why you construct $(a_n)$ to be a convergent sequence - you want to show this for all equivalent sequences, and it could be the case that both are nonconvergent. You just need them to live in $X$. You haven't proved $b \Rightarrow a$. $\endgroup$
    – Jacob
    Commented Jan 24, 2015 at 5:00
  • $\begingroup$ thanks for the correction. To prove $b\implies a$ How can I switch from sequences to functions. Should I use continuity? $\endgroup$
    – MAS
    Commented Jan 24, 2015 at 5:21
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    $\begingroup$ Your definition of equivalent sequences $(a_n)_{n\geq1}$ and $(b_n)_{n\geq1}$ simply says that the difference $a_n-b_n$ tends to $0$ (no matter whether or not each single sequence converges). $\endgroup$ Commented Jan 24, 2015 at 5:46
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    $\begingroup$ Try proving the contrapositive: Assume the negation of uniform continuity: that there exists an $\epsilon$ such that for all $\delta$, there exists points $x_\delta$ and $y_\delta$ that are $\delta$-close but $f(x_\delta)$ and $f(y_\delta)$ are $\epsilon$-apart. Then find sequences that violate (b): equivalent sequences whose images under $f$ aren't equivalent. $\endgroup$
    – Jacob
    Commented Jan 24, 2015 at 7:01

1 Answer 1

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Some remarks concerning your proof:

  • You write:

    Since $x \in X$ is an adherent point, then there exists a sequence $(a_n)$ such that $a_n \in X$ and converges to $x$.

    Note that, by definition, we have to consider any two sequences $(a_n)_n$, $(b_n)_n$ in $X$ which are equivalent. This means in particular that $(a_n)_n$ need not be convergent and it does not suffices to prove the statement for some sequence $(a_n)_n$ (convergent to $x$), but we have to prove it for all sequences $(a_n)_n$ (equiavalent to some sequence $(b_n)_n$.)

  • You write:

    Choose $\epsilon = \delta$.

    In my oppinion, that's not a good formulation - both $\delta$ and $\epsilon$ are fixed numbers and we cannot simply set $\epsilon = \delta$. In this situation it is better two introduce a new variable in the following way:

    Since $(a_n)_n$ and $(b_n)_n$ are equivalent, $$\forall \varrho>0 \, \exists N \, \forall n \geq N: |a_n-b_n| \leq \varrho.$$ Choose $\varrho = \delta$.


(a) $\Rightarrow$ (b) (Corrected version):

Let $(a_n)_{n \in \mathbb{N}}$, $(b_n)_{n \in \mathbb{N}} \subseteq X$ be two equivalent sequences and $\epsilon>0$. Since $f$ is uniformly continuous, we can choose $\delta>0$ such that $$|x-y| < \delta \Rightarrow |f(x)-f(y)|< \epsilon. \tag{1} $$ It follows from the very definition of equivalent sequences that $$\forall \varrho>0 \, \exists N \in \mathbb{N} \, \forall n \geq N: |a_n-b_n| < \varrho.$$ Choose $\varrho = \delta$. Then $|a_n-b_n| < \delta$ implies $$|f(a_n)-f(b_n)| < \epsilon$$ by $(1)$ for all $n \geq N$. Since $\epsilon>0$ is arbitrary, this shows that $(f(a_n))_{n \in \mathbb{N}}$ and $(f(b_n))_{n \in \mathbb{N}}$ are equivalent.

(b) $\Rightarrow$ (a):

Suppose that $f$ is not uniformly continuous. Then there exists $\epsilon>0$ such that for any $\delta>0$, we can find $x,y \in X$ such that $$|x-y|< \delta \qquad \text{and} \qquad |f(x)-f(y)| \geq \epsilon.$$ Choosing $\delta = \frac{1}{n}$ for $n \in \mathbb{N}$, we find sequences $(x_n)_{n \in \mathbb{N}}$ and $(y_n)_{n \in \mathbb{N}}$ such that $$|x_n-y_n| \leq \frac{1}{n} \qquad \text{and} \qquad |f(x_n)-f(y_n)|> \epsilon.$$ Show that $(x_n)_{n \in \mathbb{N}}$ and $(y_n)_{n \in \mathbb{N}}$ are equivalent and that $(f(x_n))_{n \in \mathbb{N}}$, $(f(y_n))_{n \in \mathbb{N}}$ are not equivalent.

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