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caffeinemachine
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Let $X$ and $Y$ be any two sets. Show by using Zorn's Lemma (or anything equivalent to it) that there is an injection of one into the other.
caffeinemachine said:Let $X$ and $Y$ be any two sets. Show by using Zorn's Lemma (or anything equivalent to it) that there is an injection of one into the other.
Let $\mathcal{F}$ be the set of bijective maps of the form $f:D(f)\to R(f)$ with domain $D(f)\subseteq X$ and range $R(f)\subseteq Y$, made into a partially ordered set by the relation $f\prec g\; \Longleftrightarrow\; \{D(f)\subset D(g) \text{ and } g(x) = f(x)\; \forall x\in D(f)\}$. Show that $\mathcal{F}$ satisfies the conditions for Zorn's Lemma, and then check that a maximal element is either an injection from $X$ to $Y$ or the inverse of an injection from $Y$ to $X$.caffeinemachine said:Let $X$ and $Y$ be any two sets. Show by using Zorn's Lemma (or anything equivalent to it) that there is an injection of one into the other.
I like Serena said:Erm... suppose we pick X={1,2} and Y={1}, then there is no injection...?
I think there is a trivial injection from $Y$ to $X$. Take $f:Y\to X$ as $f(1)=1$.I like Serena said:Erm... suppose we pick X={1,2} and Y={1}, then there is no injection...?
Opalg said:Let $\mathcal{F}$ be the set of bijective maps of the form $f: D(f)\to R(f)$ with domain $D(f)\subseteq X$ and range $R(f)\subseteq Y$, made into a partially ordered set by the relation $f\prec g\; \Longleftrightarrow\; \{D(f)\subset D(g) \text{ and } g(x) = f(x)\; \forall x\in D(f)\}$. Show that $\mathcal{F}$ satisfies the conditions for Zorn's Lemma, and then check that a maximal element is either an injection from $X$ to $Y$ or the inverse of an injection from $Y$ to $X$.
I think you need to define $\mathcal C$ from the start to consist of triples $(A,B,f)$ such that $A\subseteq X$, $B\subseteq Y$ and that $f$ is an injection from $A$ onto $B$. (You will need to check that this surjective property is preserved by upper bounds of chains.) Then if $(A_0,B_0,f_0)$ satisfies $B_0=Y$, that means that $f_0$ is a bijection from $A_0$ onto $Y$, and so $f_0^{-1}$ is a bijection from $Y$ onto $A_0\subseteq X$.caffeinemachine said:$\vdots$
Now by Zorn's Lemma, $\mathcal C$ has a maximal element.
Say the maximal element is $(A_0,B_0,f_0)$. If $A_0=X$ then we are done. So assume $A_0\neq X$. Now if $B_0\neq Y$ then we can easily find an element in $\mathcal C$ which is greater than $(A_0,B_0,f_0)$ and we would run into a contradiction. Thus $B_0=Y$. From here its not easy to see that $f_0^{-1}:Y\to X$ is an injection and the proof is complete.
Opalg said:I think you need to define $\mathcal C$ from the start to consist of triples $(A,B,f)$ such that $A\subseteq X$, $B\subseteq Y$ and that $f$ is an injection from $A$ onto $B$. (You will need to check that this surjective property is preserved by upper bounds of chains.) Then if $(A_0,B_0,f_0)$ satisfies $B_0=Y$, that means that $f_0$ is a bijection from $A_0$ onto $Y$, and so $f_0^{-1}$ is a bijection from $Y$ onto $A_0\subseteq X$.
Your proof is fine – I was just tying up the loose ends where you said that you could not complete the proof.caffeinemachine said:I guess including the "onto" clause will simplify the proof a bit. I have not checked the details though.
I think my proof is correct too. Isn't it?
Also, can you give some details of your proof in which you don't require triples but only pairs which seems to be considerably simpler than mine.
Oh. Now I see. Actually this problem is there in Simmon's 'Introduction to Topology and Modern Analysis' in which the author has given a hint which just uses pairs and not triples and I still don't know how to do it on those lines.Opalg said:Your proof is fine – I was just tying up the loose ends where you said that you could not complete the proof.
I don't see what you mean by using pairs rather than triples. My proof used the triples $(f,D(f),R(f))$, just like yours. And the only reason that my proof "seems to be considerably simpler" is that I left out all the details for you to fill in, which you have done.
When we say that the cardinalities of two sets are comparable, it means that we can determine whether one set has more elements than the other or if they have the same number of elements. In other words, it is possible to establish a one-to-one correspondence between the elements of the two sets, allowing us to compare their sizes.
To compare the cardinalities of two sets, we can use a one-to-one mapping function between the elements of the sets. This means that we can pair up each element in one set with a unique element in the other set. If all elements in both sets can be paired up without any leftover elements, then the cardinalities are equal. Otherwise, if there are leftover elements in one set, then it has more elements and its cardinality is larger.
Yes, it is possible for two sets with different cardinalities to be comparable. As long as there exists a one-to-one mapping between the elements of the two sets, we can compare their sizes. For example, the set of natural numbers and the set of even numbers have different cardinalities, but we can still compare them by pairing each natural number with its corresponding even number (1 with 2, 2 with 4, 3 with 6, etc.).
Comparing the cardinalities of two sets allows us to understand the sizes of these sets and how they relate to each other. It helps us determine if one set is larger or smaller than the other, and if there are any similarities or differences between the two sets. This information can be useful in various fields such as mathematics, computer science, and statistics.
Yes, there are some limitations to comparing the cardinalities of two sets. One limitation is that we can only compare the sizes of sets that have a finite number of elements. This means that we cannot compare the cardinalities of infinite sets, such as the set of all real numbers. Additionally, comparing cardinalities does not provide any information about the elements themselves, only their quantity.