Why does $C_1\cap C_2\cap C_3\cap \ldots = \{x:x=2\}$? Before I say anything:  this question is somewhat of a follow up to this question here.
Suppose $C_1,C_2, C_3\ldots$ are sets where $C_k\supset C_{k+1}$ for $k=1,2,3,\ldots$ and $\lim_{k\rightarrow\infty} C_k= C_1\cap C_2\cap C_3\cap \ldots$  My question is different than the one linked in that $C_k$ is defined by $$C_k=\left\{x:2-\frac{1}{k}<x\leq 2\right\}.$$  I am being asked to find $\lim_{k\rightarrow\infty}C_k$.  According to my book, the solution is $\{x:x=2\}.$  This is absolutely mind boggling to me because I am failing to see the difference between this $C_k$ and the one featured in the linked question because when I compute $$\lim_{k\rightarrow\infty}(2-\frac{1}{k})=2$$ I get the same answer of $$\lim_{k\rightarrow\infty} C_k=\{x: 2<x\leq 2\}=\{\}.$$  
What am I missing here?  Is the book wrong?
 A: The book is right!
Observe that if $x=2$ then $2 \geq x > 2 - \frac{1}{k}$ so $x \in \lim\limits_{k \to +\infty} C_k$.
The point is that on limits usually you lose strict inequalities, they can also have equality then.
A: Note first, that for every $k\in\mathbb{N}$, $2-\frac{1}{k}<2$, i.e. $2-\frac{1}{k}<2\leq 2$, i.e. $2\in C_k$ for every $k\in\mathbb{N}$ as $\frac{1}{k}>0$ for all these $k$. Thus, $2\in\bigcap_{k\in\mathbb{N}}C_k$. 
To show, that for every $x\neq 2:x\not\in\bigcap_{k\in\mathbb{N}}C_k$, suppose there is such an $x$. Now, thus $x\in C_k$ for every $k\in\mathbb{N}$, i.e. first $x\leq 2$ and as $x\neq 2$ we have $x<2$. Also $2-\frac{1}{k}<x$ for every $k$. Now, as $x<2$, $2-x>0$. Take $k_0$ to be the first natural number s.t. $\frac{1}{k_0}<\frac{1}{2-x}$. We have
$$x>2-\frac{1}{k_0}>2-\frac{1}{2-x}=2-(2-x)=x$$
Contradiction. Thus $x\not\in\bigcap_{k\in\mathbb{N}}C_k$.
You may not move the limit inside the set as you see. The essential thing is the behaviour of the $C_k$ under the limit, not the condition under the limit.
A: Maybe suggestive notation helps. In this case, the limiting set appears to be
$$\lim_{k\rightarrow\infty} C_k=\{x: 2^-<x\leq 2\}$$
whereas in the other question, the limiting set results in
$$\lim_{k\rightarrow\infty} C_k=\{x: 2<x\leq 2^+\}.$$
A: Your subject line says you're looking for $C_1\cap C_2 \cap C_3 \cap \cdots,$ but the body of your posting says $\lim\limits_{k\to\infty} C_k.$
The meaning of $\lim\limits_{k\to\infty} C_k$ is not so clear. But $\bigcap\limits_{k=1}^\infty C_k$ is clear. It's just the set of all objects that are a member of every one of $C_1, C_2, C_3, \ldots.$ Since $2$ is a member of every one of those, it's a member of their intersection. And it's the only member because there is no other number that is a member of all of those sets.
I would have written $\{2\}$ rather than $\{x:x=2\}.$ The latter is needlessly complicated notation.
