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Let E be a normed space and F a Banach space. Suppose we have A ⊂ E, $x_o$ ∈ Acc(A) and f : A → F , a function uniformly continuous. Show that lim x→x0 of f(x) exists.

chadi
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  • Nothing special about Banach spaces here; the result is true whenever $E$ is a metric space and $F$ is a complete metric space. – Nate Eldredge Feb 11 '18 at 07:03

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As $x_0 \in \operatorname{Acc}(A)$, it exists a sequence $(x_n)$ in $A$ such that $\lim\limits_{n \to \infty} x_n =x_0$. As $f$ is uniformly continuous, you’ll be able to prove that the sequence $(f(x_n))$ is Cauchy. As $F$ is a Banach space, $(f(x_n))$ converges let’s say to $y_0$.

Now if you take any sequence $(x_n^\prime)$ of $A$ converging to $x_0$, $(f(x_n^\prime))$ converges to $y_0$ according to $f$ uniform continuity. Which provides the result that $\lim\limits_{x \to x_0} f(x) =y_0$.

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