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How to Show Something Not Uniform Continuous

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How to show that "Uniformly continuous implies continuous"?

Solution 1

Let $f: X \to \mathbb{R}$ be uniformly continuous. Then for any $\epsilon > 0$ given there exists $\delta > 0$ such that $x,y \in X$,

$$|x - y| < \delta \Rightarrow |f(x) - f(y)|<\epsilon$$

In particular fix $a \in X$ arbitraly, then

$$|x - a| < \delta \implies |f(x) - f(a)| < \epsilon $$

for every $a \in X$ f is continuous.

An example in which $f$ is continuous but not uniformly continuous is $f(x) = \frac{1}{x}$, where $X = \mathbb{R}^+$ can you show why?

Added: Here is a hint to start. Let $0 <\epsilon<1 $, then for every $\delta > 0$ chosen we may take a natural number $n > \frac{1}{\delta}$ and set $x = \frac{1}{n}$ and $y = \frac{1}{2n}$. What is the contradiction?

Solution 2

Uniform continuity (of $f(x)$ on $A$): For every $\epsilon>0$ there is $\delta$ such that if $|x-y|<\delta$ then $|f(x)-f(y)|<\epsilon$.

while

Continuity (of $f(x)$ at $a\in A$): For every $\epsilon>0$ there is $\delta$ such that if $|x-a|<\delta$ then $|f(x)-f(a)|<\epsilon$.

Given $\epsilon>0$ then, by uniform continuity, there is $\delta$ such that for all $x,y\in A$ such that $|x-y|<\delta$ implies that $|f(x)-f(y)|<\epsilon$. In particular if we take $y=a$ we get that for all $x\in A$ such that $|x-a|<\delta$ then necessarily $|f(x)-f(a)|<\epsilon$. This last is the definition of continuity.

The key observation is that the definition of uniform continuity becomes the definition of continuity when you look to a particular point $y=a$.

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Comments

  • Assuming that a function f is uniformly continuous, and starting from the ϵ-δ definition of continuity, how does one prove that it is also continuous on the real numbers?

    Thanks.

    • If you use the epsilon delta definition, there is only one answer: it is obvious.

    • Just cover up the "uniformly" part.

    • This question appears to be off-topic because it is too obvious to be a question

    • I just don't understand the downvotes on the answers though

    • @AaronMaroja They prefer the student to go away without an answer. Since they can't prevent an answer from being seen, they canalize their impotence in the only way they can. Still the student got his[er] answer (and would have got it even if the question had been closed) and that is enough for me.

    • @Pp.. I see, I've never thought that way. Nice.

    • @AaronMaroja: briefly put, I downvoted your answer simply because it contains wrong claims. Also as I pointed out, the correct way to put it would be to say that it is obvious

    • @Laters wrong claims, yeah sure. Not even worth to discuss.

    • @AaronMaroja: $f(x)=1/x$ is uniformly continuous on a compact interval, and as you did not specify your domain (meaning that it is not even a function) the claim is simply false, I agree "not even worth to discuss" (fundamental error)

    • @Laters Oh, so this make my answer not useful? I see. I saw your records, really impressive.

    • @AaronMaroja: I did not claim that the answer is not useful to the op, just saying that the claim you did there was simply wrong. but as you corrected the mistake, I take this as expressing agreement.

    • @Laters Seriously though, you could have just told me so in a comment. I took the function as its standard definition, forget to explicity show its domain, even though is important, I don't mind agreeing with you at all. But you as an officer of this website could take things slowly as long as you not providing the best answers either.

    • @AaronMaroja The guy just said that the "correct way to put it is that it is obvious." The guy doesn't have a clue of what a proof is. So, don't expect to learn much math from him.

    • @Pp.. I couldn't agree more. What may be obvious to him is not to others, I remember when I was first learning, nothing was obvious haha, anyways. I upvoted your answer.

    • This is the most baffling comment chain I have ever read on this site.

    • lol @Pp.. if it is not obvious to you and if you really think that this requires proof then I pity you. Also shame on you for answering this kind of question, you should feel ashamed for pointing out the obvious. the op must understand that he better reflect on the definition

  • Feel free to ask.

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Source: https://9to5science.com/how-to-show-that-quot-uniformly-continuous-implies-continuous-quot

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