Suppose that $Y \subseteq X$ and that $f:X \to Y$ is injective. Then there is a bijection from $X$ to $Y$.

My attempt:

For $B\subseteq X$, $B$ is closed $\iff$ $f[B]\subseteq B$.

The closure of $A$, to be denoted by $\overline A$, is the least closed set that contains all elements of $A$ and thus defined by $\overline A=\bigcap\{B\in\mathcal P(X)\mid A\subseteq B \text{ and }B \text{ is closed}\}$.

Let $A=X\setminus Y$. We first prove that $f[\overline A]\cup A=\overline A$.

  1. $(f[\overline A]\cup A) \subseteq \overline A$

$\overline A$ is closed $\implies$ $f[\overline A] \subseteq \overline A$. Moreover, $\overline A$ is the closure of $A \implies A \subseteq \overline A$. Thus $(f[\overline A]\cup A) \subseteq \overline A$.

  1. $\overline A\subseteq (f[\overline A]\cup A)$

For $a\in \overline A$, there are only two cases.

If $a\in A$, then it follows immediately that $a\in f[\overline A]\cup A$.

If $a\notin A$, then assume the contrary that $a\notin f[\overline A]\cup A$, then $a\notin f[\overline A]$. Let $A'=\overline A \setminus \{a\}$. Since $a\notin A$, $A\subseteq A'$. We have $f[A']=f[\overline A \setminus \{a\}]=f[\overline A] \setminus \{f(a)\}$ since $f$ is injective. Since $a\notin f[\overline A]$, $f[\overline A] \setminus \{a\}=f[\overline A]$. Thus $f[A']=(f[\overline A] \setminus \{a\}) \setminus \{f(a)\} \subseteq (\overline A \setminus \{a\}) \setminus \{f(a)\} \subseteq \overline A \setminus \{a\}$. Hence $f[A'] \subseteq A'$ and consequently $A'$ is closed. To sum up, $A\subseteq A'$ and $A'$ is closed $\implies \overline A \subseteq A' \implies \overline A \subseteq (\overline A \setminus \{a\})$. This is clearly a contradiction. Hence $a\in f[\overline A]\cup A$. It follows that $\overline A\subseteq (f[\overline A]\cup A)$.

As a result, $A = f[\overline A]\cup A$. Next we prove $X\setminus \overline A= Y\setminus f[\overline A]$.

$X\setminus \overline A=X\setminus (f[\overline A]\cup A)=(X\setminus f[\overline A])\cap(X\setminus A)=(X\setminus f[\overline A])\cap Y=Y\setminus f[\overline A]$ since $Y\subseteq X$.

We define a bijection $g:X\to Y$ by $g=f_{\restriction \overline A} \cup \operatorname{id}_{X\setminus \overline A}$.

Does this proof look fine or contain gaps? Do you have suggestions? Many thanks for your dedicated help!

  • 2
    $\begingroup$ why are you talking about topology? this is a set theory question $\endgroup$ – mathworker21 Sep 21 '18 at 9:17
  • 4
    $\begingroup$ This is basically the Schröder-Bernstein theorem. $\endgroup$ – Arthur Sep 21 '18 at 9:19
  • $\begingroup$ @mathworker21 I don't understand what you meant. In my textbook Introduction to Set Theory by Thomas Jech, he introduced these concepts. What should I do? $\endgroup$ – LE Anh Dung Sep 21 '18 at 9:20
  • 1
    $\begingroup$ You should, for one thing, not speak about closed sets, open sets and closures. Just elements, sets and function values. $\endgroup$ – Arthur Sep 21 '18 at 9:20
  • 1
    $\begingroup$ @mathworker21 and Arthur, you can use topology to solve a set theory question if you want. The first line of this question is not an observation, it is a definition (of a topology on $X$). $\endgroup$ – Mees de Vries Sep 21 '18 at 9:22

Yes, your proof is perfectly correct.

There are just two typos (that I noticed): your final string of equations should end in $Y \setminus f[\overline{A}]$, and in the middle of your paragraph you use the expression $\overline{A}\setminus \{a\}A'$, where you accidentally add $A'$.

For formatting it would help if you put some of the longer strings of equations/inclusions in displays. Also, the proof of (2) doesn't need two cases: you can go straight to "Assume towards a contradiction that $a \notin f[\overline{A}] \cup A$, so that $a \notin A$ and $a \notin f[\overline{A}]$", because that is what you do in the second case anyway.

  • $\begingroup$ I got your points. Thank you so much! $\endgroup$ – LE Anh Dung Sep 21 '18 at 9:44

This is fine; it is basically Dedekind's proof; see page 447 in his Collected Works.


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