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If $A$ is finite with $|A|=n \geq 1$ and $x \in A$, I need to prove that $A - \{x\}$ is finite with $|A-\{x\}| = n-1$, where $|\cdot|$ is the cardinality.

I know that there are proofs for $|A-\{x\}| = n-1$ which rely on the result $A = (A-\{x\})\cup \{x\} $, but I was wondering if there was a way to prove this without relying on it (as relying on it would require me to prove it first).

Instead, I was thinking of proving it the following way:

Since $|A|= n \geq 1$, $A \neq \emptyset$, so $\exists$ at least one element $x \in A$.

  1. Suppose that $A = \{x\}$, then $|A|=1$ and $A - \{x\} = \emptyset$, which is certainly finite, and $|A-\{x\}|=|\emptyset|=0 = |A|-1$.

  2. Suppose that $|A|=n>1$, $x \in A$. Then, there exists a bijection $f:A \to \mathbb{N}_{n}$.

Now, to go any further with bullet point 2, I need to define a function $g:A-\{x\} \to \mathbb{N}_{n-1}$ and show that it is a bijection, but I'm not sure what function $g$ to choose. I know that I can use $f$ to help me define $g$, but how? What choice of $g$ will allow me to prove this result this way without needing to rely on $A = (A-\{x\})\cup \{x\} $?

Thank you.

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    Hint :u can use the bijection f (say) which exists between A and $N_n$ to find the bijection g – 2017-02-07
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    @RAhu I know this, but how without appealing to that result? – 2017-02-07
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    so if f(x)=i,u can take g to be g(x) = f(x) for all x s.t. f(x)< i and g(x)=f(x)-1 for all x s.t. f(x)>i ..will be a bijection. And You do not want to use that g?U want something from scratch? – 2017-02-07
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    That $A = A \setminus \{x\} \cup \{x\}$ for any $x \in A$ is a triviality, easily shown to be true. Why avoid that? – 2017-02-07
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    @HennoBrandsma because it's not just that. If you go that route, you also need to prove a lemma that $|A \cup \{x\}| = |A|+1$. – 2017-02-07
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    That's also not too hard, it seems to me. – 2017-02-08
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    @HennoBrandsma it's not,but it's a tedious induction proof. – 2017-02-08

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Suppose that $f:A \to \Bbb N_n$ is a bijection. We can find an "obvious" formula for a bijection $\tilde f:(A - \{x\}) \to \Bbb (N_n - \{f(x)\})$. Then, find an explicit bijection $g:(\Bbb N_n - \{f(x)\}) \to \Bbb N_{n-1}$. It would follow that $g \circ \tilde f$ gives us the desired bijection from $A \setminus \{x\}$ to $\Bbb N_{n-1}$.

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    you used the Schroeder-Bernstein Theorem. But these just need to be injections, right? – 2017-02-07
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    Where did I use S-B? All of the maps constructed are bijections, at least the way I've made them. – 2017-02-07
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    Right, sorry you didn't use S-B. But, why do you need to set up a composition like that? And is there anything in addition that needs to be proved here? Just want to make sure... – 2017-02-07
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    I didn't *need* to set up a composition, but it's a handy way to break the necessary construction down. You'll need explicit formulas for $\tilde f$ and $g$, and you'll need to show that these maps are bijections. I'd assume that you can say without proof that the composition of bijections is a bijection, but if not, this is also easy to prove. – 2017-02-07
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    like RAhu's suggestion above? So, say that $f: A \to \mathbb{N}_{n}$ is defined by $f(x) = i \in \mathbb{N}_{n}$ $\forall x \in A$, where if $x,y \in A$, $x \neq y$, then $f(x) \neq f(y)$ and $\forall i \in \mathbb{N}_{n}$, $\exists x \in A$ s.t. $f(x) = i$? Is that enough? I don't need to come up with anything more specific than that? – 2017-02-07
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    That's not a *definition* of $f$; it's a description (there are several $f$ that may fit this same description). Also, the phrase "is defined by $f(x) = i \in \Bbb N_n \forall x \in A$" is non-sense. Of course, we may assume that such an $f$ exists by the definition of finite cardinality, so that's not what you need to think about. You should consider how does one **use** this $f$ to construct the necessary bijection (which for me is $\tilde f \circ g$)? **This** is the question that RAhu answers. – 2017-02-07
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    I'm so confused. I guess I don't know how to even come up with an $f$. – 2017-02-07
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    You should **not** be coming up with an $f$. You **should** be coming up with a $\tilde f$, in terms of that $f$. – 2017-02-07
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    @Omnomnomnomnom I still don't understand. If all we know about $f$ is that it's a bijection from $f:A\to \mathbb{N}_{n}$, how can I come up with $\bar{f}$ in terms of $f$? And how do I know that $\bar{f}$ is a bijection? – 2017-02-07
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    Define $$ \tilde f:(A-\{x\}) \to (\Bbb N_n - \{f(x)\}\\ \tilde f(x) = f(x) $$ Now, why is this a bijection? Injectivity is easy; all we really need to think about is surjectivity. – 2017-02-07
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    @Omnomnomnom that $\bar{f}$ is bijection has to be proved.No one is saying that is very trivial.But yes,that can be proved easily if u can use the 'bijection' f.Try it out. – 2017-02-07
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    i actually wanted to say that to @JessyunBourne because he was asking that how to know $\bar{f}$ is bijection or not?...that is not known.That has to be proved.thta's what i meant.. – 2017-02-07
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    @Omnomnomnom Okay, so say I came up with the function $f^{-1} \circ \tau \circ I$ where $I: \mathbb{N}_{n-1} \to \mathbb{N}_{n}$ where for each $j \in \mathbb{N}_{n-1}$, I(j) = j; $\tau: \mathbb{N}_{n} \to \mathbb{N}_{n}$ where $\forall k \in \mathbb{N}_{n}$, $ \tau(k) = \begin{cases} n & \text{if}\, k=f(a) \\ f(a) & \text{if}\, k=n \\ k & \text{otherwise} \end{cases}$. Now, I want to show that $f^{-1} \circ \tau \circ I: \mathbb{N}_{n-1} \to A - \{x\}$ is a bijection. It's not hard to show that each piece is an injection and a composition of injections is an injection. – 2017-02-07
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    @Omnomnomnom Now to show that it's surjective is confusing me. I need to show that for $b \in A - \{x\}$ has a preimage in $\mathbb{N}_{n-1}$. But how do I do this? I'm confusing myself... – 2017-02-07
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    @JessyunBourne I don't understand why you would do such a thing. I thought the whole point was to prove the cardinality using a bijection. – 2017-02-07
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    @Omnomnomnom It is. This is the bijection. I just need to prove that it is a bijection and proving that it is surjective is proving difficult for me. – 2017-02-07
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    In any case, pretend like the question I asked is, how do I show that $f^{-1}\circ \tau \circ I$ is surjective. – 2017-02-07
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    Also, you technically should rename $f$, since you've changed the domain/codomain. That's why I renamed mine to $\tilde f$. In any case, it suffices to show that for every $y \in A$ for which $y \neq x$, there exists an $i \in \Bbb N_{n-1}$ with $f^{-1}(I(\tau(i)))) = y$. – 2017-02-07
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    @JessyunBourne I'm not planning on answering that. You should probably ask a new question at this point. – 2017-02-07