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

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!

• why are you talking about topology? this is a set theory question – mathworker21 Sep 21 '18 at 9:17
• This is basically the Schröder-Bernstein theorem. – Arthur Sep 21 '18 at 9:19
• @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? – LE Anh Dung Sep 21 '18 at 9:20
• You should, for one thing, not speak about closed sets, open sets and closures. Just elements, sets and function values. – Arthur Sep 21 '18 at 9:20
• @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 $X$). – Mees de Vries Sep 21 '18 at 9:22

## 2 Answers

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.

• I got your points. Thank you so much! – 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.