Using Schröder-Bernstein theorem to show same cardinality

Question

Use the Schröder-Bernstein theorem to show that $(0,1)\subseteq \Bbb R$ and $[0,1]\subseteq\Bbb R$ have the same cardinality.

Firstly I'm not even entirely sure about what the syntax even means. The elements of subset (0,1) and [0,1] are also included in the set of all real numbers? And [0,1] doesn't include 0 and 1 but (0,1) does? Even when I understand the syntax I probably won't know how to solve it

Answer 1

$(0,1) \subset [0,1]$ so $|(0,1)| \le |[0,1]|$.

Using the bijection $\phi(x) = \frac{1}{4}+\frac{1}{2} x$, we see that $|[0,1]| = |[\frac{1}{4},\frac{3}{4}]|$, and since $[\frac{1}{4},\frac{3}{4}] \subset (0,1)$ we have $|[0,1]| = |[\frac{1}{4},\frac{3}{4}]| \le |(0,1)|$.

Hence, using the Bieber-Cantor–Bernstein–Schröder-Heimlich-Presley theorem we have $|(0,1)| =|[0,1]|$.

Answer 2

They intend to indicate that $(0,1)$ and $[0,1]$ are meant to indicate intervals of the real line. Then $$(0,1)=\{x\in\Bbb R\mid 0<x<1\}$$ and $$[0,1]=\{x\in\Bbb R\mid0\le x\le1\}.$$

If they had said instead that $(0,1)\subseteq\Bbb Q$ and $[0,1]\subseteq\Bbb Q,$ for example, then they would be talking about rational intervals, instead--that is:

• $(0,1)=\{x\in\Bbb Q\mid 0<x<1\}$
• $[0,1]=\{x\in\Bbb Q\mid0\le x\le1\}$

What you need to do is demonstrate a one-to-one function $(0,1)\to[0,1]$ and a one-to-one function $[0,1]\to(0,1).$ The first of these tasks is very straightforward, and the second is only slightly less so. Once these tasks are done, Schröder-Bernstein Theorem tells us that the two intervals have the same cardinality.

Answer 3

First you need to understand the notation, $(a,b)=\{x\in\Bbb R\mid a<x<b\}$ and $[a,b]=\{x\in\Bbb R\mid a\leq x\leq b\}$.

Then you need to understand the problem. You need to use the Cantor-Bernstein theorem in order to show that there exists a bijection between $[0,1]$ and $(0,1)$.

Then you need to understand what is the Cantor-Bernstein theorem. The theorem states: If there are injective functions $f\colon A\to B$ and $g\colon B\to A$ then there exists a bijection $h\colon A\to B$.

So to use it you need to exhibit two injections. One injection from $[0,1]$ into $(0,1)$ and one from $(0,1)$ into $[0,1]$. I will leave this part to you.