I have a theorem in my book which says if $(f_n)$ is a sequence of functions uniformly converging on $A$ to $f$, and is continuous at some point $c \in A$, then $f$ is also continuous at this point $c$.

Does this also hold for the case in the title of this question?

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    $\begingroup$ No. For example: $f_n(x) = 0$ for all $x\not= 0$ and $f_n(0) = \frac{1}{n}$. $\endgroup$ – Winther Jun 20 '15 at 15:23

This is not true. You can have a sequence of discontinuous functions which converge uniformly to a continuous function. Consider the functions $f_n : \mathbb{R} \to \mathbb{R}$, $$f_n(x) = \begin{cases} \frac{1}{n} & x \geq 0\\ 0 & x < 0.\end{cases}$$

Note that each $f_n$ is discontinuous at zero, but $f_n$ converges uniformly to the constant zero function (which is continuous).

In fact, one can construct a family of functions $f_n$ which are nowhere continuous, yet converge uniformly to a continuous function. For example, $f_n = \frac{1}{n}\chi_{\mathbb{Q}}$.


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