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Let $a,b,c \in \mathbb{R}$ such that $a < b < c$. Define the sets: $$A := [a,b], B := [a,b] \cup \{c\}.$$I claim that there exists a bijection $f: A \to B$, i.e.,that $A$ and $B$ have the same cardinality.

To proof this, I choose $\varepsilon \in (0, b-a)$ and define the function $f:[a,b] \to [a,b] \cup \{c\}$ as follows:

$$f(x) = \begin{cases} \text{$c$}, & \text{if $x = a$} \\ \text{$x - \varepsilon$} & \text{if $x \in [a + \varepsilon,b]$} \end{cases}$$

My Idea is to "send" a to c, and shift all other values in $[a +\varepsilon, b]$ to left by $\varepsilon$, effectively filling the gap at $a$.

Now I’m asking: Is this a valid bijection, and does this argument correctly show that $A$ and $B$ are equinumerous?

I know there are many existing solutions to similar problems, but I’m not looking for a new one — rather, I’d like verification that my specific approach is correct.

Thank you in advance and best regards!

spaceisdarkgreen
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checkchecker
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    Your function is not defined for $x \in (a, a+\epsilon)$. – Li Kwok Keung Jun 11 '25 at 09:44
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    ... and it does not attain any values in $(b-\epsilon,b]$ – Reinhard Meier Jun 11 '25 at 09:50
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    Idea: Create an infinite but countable subset of $[a,b]$, let's say $S={x_0, x_1, x_2,\ldots}$ Then set $f(x_0)=c$ and $f(x_{n+1}) = x_n$ for the elements in $S$ and $f(x)=x$ if $x\not\in S$. Then show that it is a bijection – Reinhard Meier Jun 11 '25 at 09:54
  • I needed some time to think about your comments. I now understand that my current approach is meaningless. Regarding your idea: this sequence only covers some elements of $B$, right? So I would need to map the $x$ not in $S$ to the rest of the elements in $B$, correct? – checkchecker Jun 11 '25 at 10:03
  • @checkchecker yes. simply map the rest of the elements to themselves – Aniruda Suswaram Jun 11 '25 at 11:12
  • Is ${x_n} : = a + \frac{b-a}{n}$ for all $ n \in \mathbb{N}$ such a sequence? – checkchecker Jun 11 '25 at 11:40

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