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The solution which we got for the question is using Matric space and Heine Cantor theorem, which I have not studied yet. Is there any easier way to prove the question, If so please provide the way forward.
Lastly - If we prove that $F$ is uniformly convergent on every bounded interval of R, then can we say $F$ is uniformly continuous over R. please help , thank you.

Abhishek Verma
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  • Real valued continuous functions defined on a compact set are uniformly continuous. Are you familiar with the concept of “compact”? – fantasie Dec 11 '19 at 06:19
  • are you talking about - a continuous function over a closed interval is uniformly continuous. then how should we generalize this over R – Abhishek Verma Dec 11 '19 at 06:24

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Assuming that every bounded sequence of real numbers has convergent subsequence we can argue as follows:

Let $f:\mathbb R \to \mathbb R$ be continuous. Suppose $f$ is not uniformly continuous on $[a,b]$ . Then there exists $\epsilon >0$ such that for any $\delta >0$ we can find points $x,y$ with $|x-y| <\delta$ but $|f(x)-f(y)| \geq \epsilon$. Taking $\delta =\frac 1 n$ we get points $x_n,y_n$ such that $|x_n-y_n| <\frac 1 n$ but $|f(x_n)-f(y_n)| \geq \epsilon$ $\cdots$ (1) for all $n$. Now there is subsequence $(x_{n_k})$ converging to some number $x$. The inequality $|x_n-y_n| <\frac 1 n$ now shows that $y_n \to x$. Now you get a contradiction to continuity of $f$ at $x$ if you let $n \to \infty$ in (1).

By restriction $f$ is also uniformly continuous on any bounded interval, closed or not.

The second part is not true. $f(x)=x^{2}$ is uniformly continuous on every bounded interval but not on the whole line.

Kavi Rama Murthy
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