At first thought it seems obvious that they are not equinumerous since ℝ-{1} contains one less element than ℝ. However since we are dealing with infinite, uncountable sets, does this logic not hold?
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Exactly; both are uncountable and they have the same "power" : $2^{\aleph_0}$. – Mauro ALLEGRANZA Oct 10 '19 at 15:08
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So they are equinumerous? I haven’t seen that symbol before. – Partey5 Oct 10 '19 at 15:09
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It's not exactly the same question, but you can proceed similarly as here: [How to define a bijection between $(0,1)$ and $(0,1]$?](https://math.stackexchange.com/q/160738) – Martin Sleziak Nov 18 '19 at 14:07
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Yes, they are equinumerous. In general, removing a finitely many elements from an infinite set does not change its cardinality.
In this case, if you have the power of the Cantor—Schröder–Bernstein theorem, you can prove that they're equinumerous fairly easily by noting that the functions $f : \mathbb{R} - \{ 1 \} \to \mathbb{R}$ and $g : \mathbb{R} \to \mathbb{R} - \{ 1 \}$ given by $f(x) = x$ and $g(x) = 1+e^x$ are both injective.
If you can only prove equinumerosity by finding a bijection, then you'll need another trick. There is no continuous bijection $\mathbb{R} - \{ 1 \} \to \mathbb{R}$, but there are piecewise functions that work, for example $$x \mapsto \begin{cases} x & \text{if } x \not\in \mathbb{Z} \text{ or } x < 2 \\ x-1 & \text{if } x \in \mathbb{Z} \text{ and } x \ge 2 \end{cases}$$ [Of course you'd need to verify that this function does define a bijection.]
Clive Newstead
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That makes a lot of sense. I guess finding an uncountable set that was a union of sets dominated by the set wasn’t as easy as writing ℝ{1}U{1} = ℝ – Partey5 Oct 10 '19 at 15:15
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If a set $S$ has a subset $T$ where $T$ is equinumerous to $\Bbb N,$ and $s\in S,$ then $S$ is equinumerous to $S \setminus \{s\}.$
Let $T=\{t_n: n\in \Bbb N\}$ where $t_n\ne t_m$ whenever $n\ne m.$
For $x\in S,$ if $x\ne s$ and $x\not \in T,$ let $F(x)=x.$
If $s \not \in T$ let $F(s)=t_1$ and let $F(t_n)=t_{n+1}$ for each $n\in \Bbb N.$
If $s\in T$ there is a unique $n_0\in \Bbb N$ such that $s=t_{n_0},$ so let $F(t_n)=t_n$ if $n_0>n\in \Bbb N$ and let $F(t_n)=t_{n+1}$ if $n_0\le n\in \Bbb N.$
Then $F: S\to S$ \ $\{s\}$ is a bijection.
In your Q, let $S=\Bbb R$ and $s=1$ and $T=\Bbb N$ with each $t_n=n.$ So $s=t_1\in T.$ We have $F(x)=x$ if $x\in \Bbb R \setminus \Bbb N.$ And $F(x)=x+1$ if $x\in \Bbb N.$
DanielWainfleet
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Does this mean that an uncountable set A cannot be formed from the union of strictly dominated subsets Ai? My question was essentially the only idea I had that would make it true. But now that you have shown my logic was wrong it seems that it isn’t true, – Partey5 Oct 11 '19 at 10:41
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X is strictly dominated by Y if there exists an injection from X to Y but X is not equinumerous to Y. I.e. Injective but not not bijective – Partey5 Oct 11 '19 at 12:52
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If $A$ has more than one member then ${x}$ is strictly dominated by $A$ for each $x\in A,$ and $A=\cup_{x\in A}{x}$.... There $is$ an uncountable set $A$ which is the union of a countable family $F$ of subsets, and each member of $F$ is strictly dominated by $A$. Reference topics: Singular cardinals, regular carfinals . – DanielWainfleet Oct 11 '19 at 13:14