Cardinality of $[0,1]$

Question

Assuming we know that the cardinality of $(0,1)$ is $c$. Can we from this determine the cardinality of $[0,1]$? Is it valid to say that $[0,1]$ also has cardinality c because adding a finite number of elements to $(0,1)$ (i.e. including the boundaries) does not affect the cardinality?

Answer 1

Yes, it is correct that $[0,1]$ has cardinality $\mathfrak{c}$; however, if it isn't clear to you, then you shouldn't think of it as "because adding finitely many points doesn't affect cardinality". Think of it in terms of bijections - $[0,1]$ will have cardinality $\mathfrak{c}$ if and only if there is a bijection between $(0,1)$ and $[0,1]$. Or, if you know the Cantor-Bernstein (Schroder-Bernstein, etc.) theorem, then you can just look for an injection from $(0,1)$ into $[0,1]$ and another from $[0,1]$ into $(0,1)$.

For example, we know $\mathbb{N}$ has cardinality $\aleph_0$. Say we want to know whether $\{0.5\} \cup \mathbb{N}$ (that is, the natural numbers with $0.5$ thrown in) also has cardinality $\aleph_0$. To do this, we establish a bijection $f:\{0.5\}\cup\mathbb{N} \to \mathbb{N}$. The most natural approach (in my opinion) would be to say $f(0) = 0$, $f(n) = n + 1$ for all whole numbers $n > 0$, and $f(0.5) = 1$. This is a bijection: it sends every member of $\{0.5\}\cup\mathbb{N}$ to a different member of $\mathbb{N}$, and every member of $\mathbb{N}$ is hit by something from $\{0.5\}\cup\mathbb{N}$. So $|\{0.5\}\cup\mathbb{N}| = |\mathbb{N}| = \aleph_0$.

You should do something similar to see that $[0,1]$ has the same cardinality as $(0,1)$; I'll leave it to you to find a similar argument.

Answer 2

Define $f:(0,1)\to [0,1]$ as follow \begin{align*} f(x)=\begin{cases} 0&, \quad x=\frac{1}{2}\\ 1&, \quad x=\frac{1}{3}\\ \frac{1}{n-2}&, \quad x=\frac{1}{n}\quad,\quad n\ge 4\\ x&, \quad x\ne\frac{1}{n}\\ \end{cases} \end{align*} Note $f$ is an bijection function from $(0,1) $ to $[0,1]$.

Answer 3

It's not necessary to find a bijection to answer your question.

If we know that there are inclusions (injections),

$$(0,1) \subset [0,1] \subset \mathbb{R}$$

then we have that

$$card(0,1) \leq card[0,1] \leq card\mathbb{R}$$

If you already know $card(0,1) = card\mathbb{R}$ then...