let $\Lambda= (1,10]$ and for each $\alpha \in \Lambda$ let $A_\alpha =\{0, \frac 1 \alpha\}$ let $\Lambda=(1,10]$ and for each $\alpha \in \Lambda $ let $A_\alpha=\{0,\frac{1}{\alpha}\}$
find $\bigcup \{A_\alpha: \alpha \in  \Lambda\}$
I know $\bigcup \{A_\alpha :\alpha \in  \Lambda\}=\{1,\frac 1 2,\ldots\}$
but I do not know how the final answer can be , please any help 
 A: The answer is $[0.1,1)\cup{0}$ $\alpha$ varies from $1$(exclusive) to $10$. So $1/\alpha$ varies from $0.1$ to $1$(excluding) and of course you have 0.
A: First, note that $1$ is not a member of $\displaystyle\bigcup\{ A_\alpha : 1<\alpha\le 10\}.$ For anything to be a member of the union, it must be a member of at least one of the sets $A_\alpha$, and $1$ is not a member of any of those since the definition specified $1<\alpha$ rather than $1\le\alpha.$
Your suggestion that the union is $\{1,\frac 1 2, \ldots\}$ seems to suggest that nothing between $1$ and $1/2$ is included. But if $(1,10]$ includes all real numbers bigger than $1$ but not bigger than $10,$ then, for example $1.01\in(1,10],$ so $1/1.01$ is a member of the union.
If $1 < \alpha \le 10$ then $1 > \dfrac 1 \alpha \ge \dfrac 1 {10}.$ That means every member of the union except $0$ is a member of $[1/10,1)$, so $\bigcup\{A_\alpha : \alpha \in (1,10]\} \subseteq [1/10,1) \cup\{0\}.$
After that it is enough to show the inverse inclusion, i.e. $\displaystyle \bigcup \{ A_\alpha : \alpha\in(1,10]\} \supseteq [1/10,\ 1) \cup\{0\}.$ So suppose $\beta$ is any member of $[1/10,\ 1).$ Then $1/\beta\in(1,10].$ So let $\alpha = 1/\beta$ and then you have $\beta = 1/\alpha\in \{0,1/\alpha\} = A_\alpha,$ so $\beta$ is a member of the union.
A: Here is how I would calculate the answer.  Expanding the definitions for $\;\Lambda\;$ and $\;A_\alpha\;$, you are asked to simplify
$$
\bigcup\left\{ \left\{ 0 , \tfrac 1 \alpha \right\} \mid \alpha \in (1,10] \right\}
$$
So let's just calculate which $\;x\;$ are in that set, by expanding definitions and simplifying.$%
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\Ref}[1]{\text{(#1)}}
\newcommand{\then}{\Rightarrow}
\newcommand{\when}{\Leftarrow}
%$
$$\calc
    x \in \bigcup\left\{ \left\{ 0 , \tfrac 1 \alpha \right\} \mid \alpha \in (1,10] \right\}
\op\equiv\hint{basic property of $\;\bigcup\;$}
    \langle \exists \alpha :: \alpha \in (1,10] \;\land\; x \in  \left\{ 0 , \tfrac 1 \alpha \right\} \rangle
\op\equiv\hint{expand set notations}
    \langle \exists \alpha :: 1 \lt \alpha \le 10 \;\land\; (x = 0 \lor x = \tfrac 1 \alpha) \rangle
\op\equiv\hints{logic: extract part not using $\;\alpha\;$ out of $\;\exists \alpha\;$,}\hint{using $\;\langle \exists \alpha :: 1 \lt \alpha \le 10 \rangle\;$ -- to simplify}
    x = 0 \;\lor\; \langle \exists \alpha :: 1 \lt \alpha \le 10 \;\land\; x = \tfrac 1 \alpha \rangle
\op\equiv\hint{arithmetic, using $\;x \ne 0\;$-- to prepare for one-point rule}
    x = 0 \;\lor\; \langle \exists \alpha :: 1 \lt \alpha \le 10 \;\land\; \alpha = \tfrac 1 x \rangle
\op\equiv\hint{one-point rule}
    x = 0 \;\lor\; 1 \lt \tfrac 1 x \le 10
\op\equiv\hint{arithmetic}
    x = 0 \;\lor\; \tfrac 1 {10} \le x \lt 1
\op\equiv\hint{introduce set notiations}
    x \in \left\{ 0 \right\} \cup [\tfrac 1 {10}, 1)
\endcalc$$
And by set extensionality, that proves
$$
\bigcup\left\{ \left\{ 0 , \tfrac 1 \alpha \right\} \mid \alpha \in (1,10] \right\} \;=\; \left\{ 0 \right\} \cup [\tfrac 1 {10}, 1)
$$
A: The answer is the set $[1/10,1)\cup \{0\}$.
