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I have to prove that if $A$ is infinite and $B$ countable, then there is an injection from $A+B$ to $A$. Here, $A+B = \{(0,a)\mid a \in A \} \cup \{ (1,b) \mid b \in B \}$. I have to use that if $X$ is infinite, there is an injective function from $ \Bbb N$ to $X$. Countable is defined as: there is a surjection from $\Bbb N$ to $B$.

Vivaan Daga
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anoniem
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  • Not at all, because I don't understand the interlacing and it doesn't use the statement 'if X is infinite, there is an injective function from N to X', which I have to use – anoniem Mar 26 '23 at 09:31
  • I suggest you edit your answer to indicate what exactly you don’t understand about the other answer. Also please use MathJax, you can see how @RobertShore formatted your answer by clicking the “edit” button. – Vivaan Daga Mar 26 '23 at 09:40
  • It's just another question. I have to use 'if X is infinite, there is an injective function from N to X' – anoniem Mar 26 '23 at 09:42
  • Here is a MathJax tutorial: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Robert Shore Mar 26 '23 at 09:42
  • Thanks. Can anyone help me please? – anoniem Mar 26 '23 at 09:44
  • "If $X$ is infinite, then there is an injective function from $\mathbb{N}$ to $X$" = "We can select an enumeration of a countable subset $S$ of $X$"; namely, let $S$ be the image of $\mathbb{N}$ under the injective function. – Patrick Stevens Mar 26 '23 at 09:44
  • That's true. But then? – anoniem Mar 26 '23 at 09:47
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    Then you follow the answer in that duplicate. Please rephrase your question to identify what precisely you don't understand about the answer. – Patrick Stevens Mar 26 '23 at 09:47
  • I don't understand the interlacing they are talking about. – anoniem Mar 26 '23 at 09:48
  • https://math.stackexchange.com/questions/4071441/let-a-b-be-some-sets-such-that-a-setminus-b-is-infinite-while-b-countable?rq=1 is also a duplicate, by the way, although the set you've denoted as $A$ is what they've denoted as $A \setminus B$ (so their $A$ is your $A + B$). – Patrick Stevens Mar 26 '23 at 09:50

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Your question appears to be: how does the "interlacing" work in an existing answer?

To enumerate the disjoint union $A + B$ given an enumeration of $A$ and an enumeration of $B$, simply hop between the enumerations: $a_1, b_1, a_2, b_2, \dots$. In that answer, they've ensured that it's a disjoint union by additionally tagging each element with the set it came from: $a_i$ is tagged with a $0$, and $b_i$ is tagged with a $1$, so they've constructed the sequence $(0, a_1), (1, b_1), (0, a_2), (1, b_2), \dots$. This ensures that the sequence contains no duplicate elements that might arise if $A$ and $B$ intersected nontrivially. (This is a standard trick. Another way to solve the problem would simply be to say "use the sequence $a_1, b_1, a_2, b_2, \dots$, but omit any duplicates".)

Patrick Stevens
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