Prove that a nonempty set ${T_1}$ is finite if and only if there is a bijection from ${T_1}$ onto a finite set ${T_2}$.

Now,I think my proof is correct but I want the opinion of other people because I have started analysis. Please help me perfect my thought process and proof writing by checking my solution.

${{Proof : }}$ Since ${T_2}$ is a finite set, so there is a bijection from ${T_2}$ onto ${\mathbb{N}_m}$ for some ${m\in\mathbb{N}}$. Let,

${f : T_2\rightarrow{N}_m}$ _(1)

be a bijection from ${T_2}$ to ${\mathbb{N}_m}$. Now, if ${T_1}$ is a bijection onto ${T_2}$, then

${g : T_1{\rightarrow}T_2}$. _(2)

Now, from ${(1)}$ & ${(2)}$,

${f(g(T_1)) = {\mathbb{N}_m}}$.

So, ${f.o.g : T_1\rightarrow\mathbb{N}_m}$ is a bijection from ${T_1}$ onto ${{\mathbb{N}_m}}$. Therefore ${T_1}$ is a finite set.

Please correct me if my proof is wrong or if it needs some more argument. Thank you! :))


When you see a 'if and only if', you have two directions to prove (the if, and the only if). What you have is about the point that if $T_1$ has a bijection onto some finite set $T_2$, then $T_1$ is also finite. But you must also show the converse, that if $T_1$ is finite then there is some finite set $T_2$ for which $T_1$ has a bijection onto $T_2$. This direction is trivial, as you can take $T_2 = T_1$ and there certainly is a bijection between $T_1$ and $T_1$. But it needs to be stated.

As for your proof, what you end up with is that $T_1$ has a bijection onto some finite set $\mathbb{N}_m$. But we already knew that, since there is a bijection between $T_1$ and finite $T_2$. So we gained no new information. Actually, you could set $\mathbb{N}_m = T_2$ and nothing new was deduced. What you can look into is the condition on the cardinalities of $T_1$ and $T_2$ for them to admit a bijection.

  • $\begingroup$ Thank you! Can you please show me the correct proof so that I can understand how to proceed in similar situations? Thank you... :) $\endgroup$ – Abir Mukherjee Apr 25 '14 at 19:47
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    $\begingroup$ Well, maybe you should try first :) Can you find an argument using the cardinalities of $T_1$ and $T_2$ if they admit a bijection ? $\endgroup$ – Manuel Lafond Apr 25 '14 at 19:52
  • $\begingroup$ I think the cardinalities of ${T_1}$ and ${T_2}$ must be same since they are in bijection. If ${T_1}$ is not a finite set then ${|T_1|>|T_2|}$, which contradicts the fact that they are in bijection. Thus ${T_1}$ is a finite set.Am I right? :-) $\endgroup$ – Abir Mukherjee Apr 25 '14 at 21:54
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    $\begingroup$ That is correct and convincing enough for me :) Now, you might want to convince yourself that the cardinalities must be the same. Essentially, a bijection between $X$ and $Y$ maps each element of $X$ to a distinct element of $Y$, such that every element of $Y$ got mapped to some element - which is impossible if $|X| \neq |Y|$. This can be formalized using the definition of a bijection : it is a surjective and injective function. $\endgroup$ – Manuel Lafond Apr 26 '14 at 4:51

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