Prove that a nonempty set $T_1$ is finite if and only if there is a bijection from $T_1$ onto a finite set...
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Prove that a nonempty set $T_1$ is finite if and only if there is a bijection from $T_1$ onto a finite set $T_2.$
proof. ($leftarrow)$ Let $h:T_2 rightarrow T_1$ be a bijection from $T_2$ onto $T_1$. And let $f:T_2 rightarrow mathbb{N_m}$ be a bijection from $T_2$ onto $mathbb{N_m}$, where $m in mathbb{N}$. Since $h$ is a bijective function, we can suppose that $h$ is the composite function $(g circ f)(x)$, where $x$ is an element of the nonempty set $T_2$, and $g:mathbb{N_m} rightarrow T_1$ is a bijective function from $mathbb{N_m}$ onto $T_1$. Since $T_1$ has a one-to-one correspondence with $mathbb{N_m}$, it follows that $T_1$ is finite. QED
It seems that I've proved the ($leftarrow)$ direction. For the ($rightarrow)$ part, do I just show that the finite set $T_1$ is in a one-to-one correspondence with $mathbb{N_n}$, for some $n in mathbb{N}$. Then, I show that $mathbb{N_n}$ is in a one-to-one correspondence with an arbitrary set $T_2$. And then I form a composite function to show that there's a bijection between $T_1$ and $T_2$? I feel like I'm making too many assumptions for both sides of the proof.
real-analysis
asked Nov 17 at 0:22
K.M
629312
add a comment |
up vote
0
down vote
favorite
Prove that a nonempty set $T_1$ is finite if and only if there is a bijection from $T_1$ onto a finite set $T_2.$
proof. ($leftarrow)$ Let $h:T_2 rightarrow T_1$ be a bijection from $T_2$ onto $T_1$. And let $f:T_2 rightarrow mathbb{N_m}$ be a bijection from $T_2$ onto $mathbb{N_m}$, where $m in mathbb{N}$. Since $h$ is a bijective function, we can suppose that $h$ is the composite function $(g circ f)(x)$, where $x$ is an element of the nonempty set $T_2$, and $g:mathbb{N_m} rightarrow T_1$ is a bijective function from $mathbb{N_m}$ onto $T_1$. Since $T_1$ has a one-to-one correspondence with $mathbb{N_m}$, it follows that $T_1$ is finite. QED
It seems that I've proved the ($leftarrow)$ direction. For the ($rightarrow)$ part, do I just show that the finite set $T_1$ is in a one-to-one correspondence with $mathbb{N_n}$, for some $n in mathbb{N}$. Then, I show that $mathbb{N_n}$ is in a one-to-one correspondence with an arbitrary set $T_2$. And then I form a composite function to show that there's a bijection between $T_1$ and $T_2$? I feel like I'm making too many assumptions for both sides of the proof.
real-analysis
asked Nov 17 at 0:22
K.M
629312
1
You don't need an arbitrary $T_2$, you just need to find one that works. Along those lines, $mathbb{N}_n$ is a finite set, and showing that the finite set $T_1$ is in one-to-one correspondence with $mathbb{N}_n$ is the same as showing that a bijection exists from $T_1$ to $mathbb{N}_n$.
– Tartaglia's Stutter
Nov 17 at 0:47
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Prove that a nonempty set $T_1$ is finite if and only if there is a bijection from $T_1$ onto a finite set $T_2.$
proof. ($leftarrow)$ Let $h:T_2 rightarrow T_1$ be a bijection from $T_2$ onto $T_1$. And let $f:T_2 rightarrow mathbb{N_m}$ be a bijection from $T_2$ onto $mathbb{N_m}$, where $m in mathbb{N}$. Since $h$ is a bijective function, we can suppose that $h$ is the composite function $(g circ f)(x)$, where $x$ is an element of the nonempty set $T_2$, and $g:mathbb{N_m} rightarrow T_1$ is a bijective function from $mathbb{N_m}$ onto $T_1$. Since $T_1$ has a one-to-one correspondence with $mathbb{N_m}$, it follows that $T_1$ is finite. QED
It seems that I've proved the ($leftarrow)$ direction. For the ($rightarrow)$ part, do I just show that the finite set $T_1$ is in a one-to-one correspondence with $mathbb{N_n}$, for some $n in mathbb{N}$. Then, I show that $mathbb{N_n}$ is in a one-to-one correspondence with an arbitrary set $T_2$. And then I form a composite function to show that there's a bijection between $T_1$ and $T_2$? I feel like I'm making too many assumptions for both sides of the proof.
real-analysis
asked Nov 17 at 0:22
K.M
629312
Prove that a nonempty set $T_1$ is finite if and only if there is a bijection from $T_1$ onto a finite set $T_2.$
proof. ($leftarrow)$ Let $h:T_2 rightarrow T_1$ be a bijection from $T_2$ onto $T_1$. And let $f:T_2 rightarrow mathbb{N_m}$ be a bijection from $T_2$ onto $mathbb{N_m}$, where $m in mathbb{N}$. Since $h$ is a bijective function, we can suppose that $h$ is the composite function $(g circ f)(x)$, where $x$ is an element of the nonempty set $T_2$, and $g:mathbb{N_m} rightarrow T_1$ is a bijective function from $mathbb{N_m}$ onto $T_1$. Since $T_1$ has a one-to-one correspondence with $mathbb{N_m}$, it follows that $T_1$ is finite. QED
It seems that I've proved the ($leftarrow)$ direction. For the ($rightarrow)$ part, do I just show that the finite set $T_1$ is in a one-to-one correspondence with $mathbb{N_n}$, for some $n in mathbb{N}$. Then, I show that $mathbb{N_n}$ is in a one-to-one correspondence with an arbitrary set $T_2$. And then I form a composite function to show that there's a bijection between $T_1$ and $T_2$? I feel like I'm making too many assumptions for both sides of the proof.
real-analysis
real-analysis
asked Nov 17 at 0:22
K.M
629312
asked Nov 17 at 0:22
K.M
629312
asked Nov 17 at 0:22
K.M
629312
asked Nov 17 at 0:22
K.M
629312
asked Nov 17 at 0:22
K.M
629312
629312
1
You don't need an arbitrary $T_2$, you just need to find one that works. Along those lines, $mathbb{N}_n$ is a finite set, and showing that the finite set $T_1$ is in one-to-one correspondence with $mathbb{N}_n$ is the same as showing that a bijection exists from $T_1$ to $mathbb{N}_n$.
– Tartaglia's Stutter
Nov 17 at 0:47
add a comment |
1
You don't need an arbitrary $T_2$, you just need to find one that works. Along those lines, $mathbb{N}_n$ is a finite set, and showing that the finite set $T_1$ is in one-to-one correspondence with $mathbb{N}_n$ is the same as showing that a bijection exists from $T_1$ to $mathbb{N}_n$.
– Tartaglia's Stutter
Nov 17 at 0:47
1
1
You don't need an arbitrary $T_2$, you just need to find one that works. Along those lines, $mathbb{N}_n$ is a finite set, and showing that the finite set $T_1$ is in one-to-one correspondence with $mathbb{N}_n$ is the same as showing that a bijection exists from $T_1$ to $mathbb{N}_n$.
– Tartaglia's Stutter
Nov 17 at 0:47
You don't need an arbitrary $T_2$, you just need to find one that works. Along those lines, $mathbb{N}_n$ is a finite set, and showing that the finite set $T_1$ is in one-to-one correspondence with $mathbb{N}_n$ is the same as showing that a bijection exists from $T_1$ to $mathbb{N}_n$.
– Tartaglia's Stutter
Nov 17 at 0:47
add a comment |
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1
You don't need an arbitrary $T_2$, you just need to find one that works. Along those lines, $mathbb{N}_n$ is a finite set, and showing that the finite set $T_1$ is in one-to-one correspondence with $mathbb{N}_n$ is the same as showing that a bijection exists from $T_1$ to $mathbb{N}_n$.
– Tartaglia's Stutter
Nov 17 at 0:47