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I have tried to solve the following question but I haven't gotten an answer. Show that:

  1. $\lim\ [0,1-1/n] = [0,1)$.

  2. $\lim\ [0,1-1/n) = [0,1)$.

all the limits having $n\to\infty$.

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Should the second problem be the same as the first? – AD. Nov 4 '11 at 6:12
@AD. In 1. you have open intervals while in 2. you have closed intervals on the left hand side. – t.b. Nov 4 '11 at 6:30
Both sequences of sets $(A_n)$ are increasing hence their limits are the unions of the $A_n$. Every $A_n$ is included in $A=[0,1)$ hence the two limit-sets are subsets of $A$. It remains to show that, in both cases, every $x$ in $A$ belongs to some $A_n$. – Did Nov 4 '11 at 6:37
@t.b. Ahh... I see thanks =) – AD. Nov 4 '11 at 19:51

Some hints

  1. Denote the sets of 1. by $I_n$ and those of 2. by $J_n$, then $J_n \subset I_n \subset J_{n+1}$. Hence their limits must be the same.

  2. The element $1$ does not belong to any sets.

  3. If $0\le x<1$ then $x<1-1/n$ for $n$ large enough.

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Let $J_n=[0,1-1/n)$ and $A_n=[0,1-1/n]$. In what sense do you mean $\lim\limits_{n \to \infty}J_n$ and $\lim\limits_{n \to \infty}A_n$? The lim inf and the lim sup when they coincide? If so, you have $$\liminf_{n \to \infty}\;J_n = \bigcup_{n=1}^\infty\left(\bigcap_{m=n}^\infty J_m\right)\tag{1}$$ and $$\limsup_{n \to \infty}\;J_n = \bigcap_{n=1}^\infty\left(\bigcup_{m=n}^\infty J_m\right),\tag{2}$$ and similarly for the $A_n$’s. If this is the notion of limit that you’re using, you first have to show that these are equal, since that’s what it means to say that the limit (in this sense) exists.

The intervals $J_n$ are increasing: $J_1\subseteq J_2\subseteq J_3\subseteq \dots$. Thus, $J_n\subseteq J_m$ whenever $n\le m$, and hence $$\bigcap_{m=n}^\infty J_m = J_n\;.$$ We can therefore simplify $(1)$: $$\liminf_{n\to\infty}\;J_n = \bigcup_{n=1}^\infty J_n\;.$$ Call this set $L$. Moreover, the fact that the intervals are increasing implies (why?) that $$\bigcup_{m=n}^\infty J_m=\bigcup_{m=1}^\infty J_m = L$$ for every $n$, so we can rewrite $(2)$ as $$\limsup_{n\to\infty}\;J_n = \bigcap_{n=1}^\infty L = L\;.$$ This shows that $\liminf_n J_n = \limsup_n J_n$ and justifies talking about $\lim_n J_n$: $$\lim_{n\to\infty}J_n = L\;.$$ Recall that by definition $L$ is just the union of the $J_n$. What is that union? Obviously $J_n\subseteq [0,1)$ for every $n$, so $L\subseteq [0,1)$. All you need to do to finish the argument is show that $[0,1)\subseteq L$.

HINT: What is $$\lim_{n\to\infty}\left(1-\frac1n\right)\;?$$

The other problem can be worked almost identically.

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My first thought was that your statement should have said $$ \bigcup_{n=1}^\infty \left[0,1-\frac1n\right] = [0,1), $$ which is certainly true.

But then it occurred to me that a reasonable interpretation of the limit is as the pointwise limit of the sequence of indicator functions (sometimes called "characteristic functions", but I preferred to avoid that term because in probability theory it means something quite different). Or more precisely, the set whose indicator function is that pointwise limit.

Let $$ \chi_A (x) =\begin{cases} 1 & \text{if } x\in A \\ 0 & \text{if }x\not\in A \end{cases} $$ denote the indicator function of the set $A$. Then $$ \lim_{n\to\infty}\chi_{[0,1-1/n]}(1) $$ is the limit of a sequence of $0$s, so it is $0$. If $x\in[0,1)$ then $\lim\limits_{n\to\infty}\chi_{[0,1-1/n]}(x)$ is the limit of a sequence that typically looks like this: $$ 0,0,0,0,0,\ldots\ldots,0,0,0,1,1,1,1,1,1,1,1,1\ldots $$ with all $1$s after some point. The number of $0$s depends on $x$, and is $0$ if $x=0$ and $>0$ if $0<x<1$. So its limit is $1$.

And if $x>1$ or $x<0$ then you get an infinite sequence of $0$s.

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