Question Number 227916 by Lara2440 last updated on 01/Mar/26
Answered by Kassista last updated on 01/Mar/26
![let A=[0,1] and B=[0,1]^2 there is an injection from A to B f:A→B s.t. f(x)=(x,0) pf: given a,b∈A f(a)=f(b)⇒(a,0)=(b,0)⇒a=b moreover, there is an injection from B to A ∀(x,y)∈B, take the terminating decimal expansion i.e. the one tha ends in zeros and is unique: x=x_1 x_2 x_3 ..., y=y_1 y_2 y_3 ... g: B→A s.t. g(x,y)=x_1 y_1 x_2 y_2 x_3 y_3 ..._2 pf: g(x,y)=g(x′,y′)⇒x_1 y_1 x_2 y_2 ..._2 =x′_1 y′_1 x′_2 y′_2 ..._2 { ((x_1 =x′_1 )),((y_1 =y_1 ′)),((x_2 =x′_2 )),((y_2 =y′_2 ...)) :}⇒x=x′ and y=y′ therefore, there exists a bijection between the sets, but I also struggle to show it](https://www.tinkutara.com/question/Q227920.png)
$$ \\ $$$${let}\:{A}=\left[\mathrm{0},\mathrm{1}\right]\:{and}\:{B}=\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} \\ $$$${there}\:{is}\:{an}\:{injection}\:{from}\:{A}\:{to}\:{B} \\ $$$${f}:{A}\rightarrow{B}\:{s}.{t}.\:{f}\left({x}\right)=\left({x},\mathrm{0}\right) \\ $$$${pf}:\:{given}\:{a},{b}\in{A} \\ $$$${f}\left({a}\right)={f}\left({b}\right)\Rightarrow\left({a},\mathrm{0}\right)=\left({b},\mathrm{0}\right)\Rightarrow{a}={b} \\ $$$$ \\ $$$${moreover},\:{there}\:{is}\:{an}\:{injection}\:{from}\:{B}\:{to}\:{A} \\ $$$$\forall\left({x},{y}\right)\in{B},\:{take}\:{the}\:{terminating}\:{decimal}\:{expansion} \\ $$$${i}.{e}.\:{the}\:{one}\:{tha}\:{ends}\:{in}\:{zeros}\:{and}\:{is}\:{unique}: \\ $$$${x}={x}_{\mathrm{1}} {x}_{\mathrm{2}} {x}_{\mathrm{3}} …,\:{y}={y}_{\mathrm{1}} {y}_{\mathrm{2}} {y}_{\mathrm{3}} … \\ $$$${g}:\:{B}\rightarrow{A}\:{s}.{t}.\:{g}\left({x},{y}\right)={x}_{\mathrm{1}} {y}_{\mathrm{1}} {x}_{\mathrm{2}} {y}_{\mathrm{2}} {x}_{\mathrm{3}} {y}_{\mathrm{3}} …_{\mathrm{2}} \\ $$$${pf}:\:{g}\left({x},{y}\right)={g}\left({x}',{y}'\right)\Rightarrow{x}_{\mathrm{1}} {y}_{\mathrm{1}} {x}_{\mathrm{2}} {y}_{\mathrm{2}} …_{\mathrm{2}} ={x}'_{\mathrm{1}} {y}'_{\mathrm{1}} {x}'_{\mathrm{2}} {y}'_{\mathrm{2}} …_{\mathrm{2}} \\ $$$$\begin{cases}{{x}_{\mathrm{1}} ={x}'_{\mathrm{1}} }\\{{y}_{\mathrm{1}} ={y}_{\mathrm{1}} '}\\{{x}_{\mathrm{2}} ={x}'_{\mathrm{2}} }\\{{y}_{\mathrm{2}} ={y}'_{\mathrm{2}} …}\end{cases}\Rightarrow{x}={x}'\:{and}\:{y}={y}' \\ $$$${therefore},\:{there}\:{exists}\:{a}\:{bijection}\:{between}\:{the}\:{sets}, \\ $$$${but}\:{I}\:{also}\:{struggle}\:{to}\:{show}\:{it} \\ $$$$ \\ $$
Commented by Lara2440 last updated on 02/Mar/26
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