If we let $f_n(x)= 1,$ if $x\in\{1,\frac{1}{2},...,\frac{1}{n}\}$ and $0$, otherwise. How can I find the pointwise limit $f$ of this sequence?
Pointwise limit of $f$
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convergence
piecewise-continuity
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0Clearly if $x \not \in \{\frac{1}{n} : n \in \Bbb{N}\}$ then $f_n(x) \equiv 0 \to 0$. On the other hand if $x \in \{\frac{1}{n} : n \in \Bbb{N}\}$ then $f_n(x) \equiv 1$ for all $n$ sufficiently large so $f_n(x) \to 1$. – 2017-02-09
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0what does the equal sign plus a extra line mean? – 2017-02-09
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0It means that the equality holds for all $n$ and all $x$. – 2017-02-09
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0so you must look at $f_n(x)$ for both conditions? – 2017-02-09
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0Like the answerer wrote, the way to go is to treat those two exhaustive cases, yes. – 2017-02-09
1 Answers
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Argue by cases.
If $x$ is of the form $\frac1k$ for some $k$, then convince yourself that $x\in\{1,\frac12,\ldots,\frac1n\}$ for all sufficiently large $n$, which means $f_n(x)=1$ for all sufficiently large $n$. Thus $f_n(x)\to1$.
If $x$ is not of the form $\frac1k$ for any $k$, then argue that $x$ is never a member of the set $\{1,\frac12,\ldots,\frac1n\}$ and therefore $f_n(x)=0$ for all $n$. Thus $f_n(x)\to0$.
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0@dgail This is the solution; you need to be confident that you understand the reasoning. – 2017-02-09
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0okay thank you I will try and make sense of this reasoning – 2017-02-09