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The function that I am having trouble with is: $$ { f }_{ n }(x)=\begin{cases} 1\quad ,\quad x \in\{ 1,\frac { 1 }{ 2 } ,...,\frac { 1 }{ n } \} \\ 0\quad ,\quad otherwise \end{cases} $$ and in particular I am trying to show that the each $f_{n}$ is continuous at x = 0, and also I am trying to find the pointwise limit of this function.

  • For the continuity at x=0 I believe I would have to claim that since the set: $$ {1,\frac{1}{2},...,\frac{1}{n} } $$ tends to zero as $ n \rightarrow \infty $ and that the second condition would be when x is exactly 0 it is zero because it doesn't belong in the set so it would then be continuous at x=0.
  • As for the pointwise I belive I would use the same argument to show that the pointwise limit would be 0.

If these are correct or incorrect any suggestions would be appreciated.

  • 0
    A nice follow up question would be to see if $f$ is also differentiable at $0$ – 2017-02-08

2 Answers 2

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Indeed, it is true that $f_n$ is continuous at $x=0$. Note that, $f_n(x)=0$ for $x\in(-\frac1n, \frac1n)$, that is, $f_n$ is constant in a neighbourhood of $x=0$.

Now, for the pointwise limit, note that if $m\in\mathbb N$, then $f_n(\frac1m)=1$ if $n\geq m$. For the other points, it is $f_n(x)=0 \ \forall n\in\mathbb N$.

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The continuity of $f_n(x)$ in $0$ is ensured because this function is constant in a neighbourhood of $0$ i.e. $\forall |x|<\frac{1}{n},\ f_n(x)=0$.

The pointwise limit is defined by $f(x)=\begin{cases} 1 \quad \forall x\in\{\frac 1k\,|\,k\in \mathbb N^*\}\\ 0 \quad otherwise\\ \end{cases}$

This function $f$ is not the null function, there are an infinity of points where its value is $1$.

But $f$ is not continuous in $0$, because for any $\delta>0$, let's take $n_0=\lfloor \frac 1\delta\rfloor+1$ so that $\frac 1{n_0}<\delta$

$f$ continuity would be $\forall \varepsilon>0,\ |x|<\delta\Rightarrow|f(x)<\varepsilon$ but $f(\frac 1{n_0})=1>\varepsilon$ so this is failing.