Corresponding Counts [3 points]

Prove that $|\{x\in \mathbb{R}|0\le x\le 1\}|=|\{x\in \mathbb{R}|4<x<7\}|$.

Jonathan Kent
2022-05-23
Answered

Help with using the Schroeder-Bernstein Theorem?

Corresponding Counts [3 points]

Prove that $|\{x\in \mathbb{R}|0\le x\le 1\}|=|\{x\in \mathbb{R}|4<x<7\}|$.

Corresponding Counts [3 points]

Prove that $|\{x\in \mathbb{R}|0\le x\le 1\}|=|\{x\in \mathbb{R}|4<x<7\}|$.

You can still ask an expert for help

Terrance Phillips

Answered 2022-05-24
Author has **10** answers

Step 1

The injection $x\to x+5$ shows that we have an injection from [0, 1] into (4, 7), so by definition $|[0,1]\le |(4,7)|$.

Step 2

The injection $x\to \frac{x}{7}$ shows that we have an injection from (4, 7) into [0, 1] (the image of (4, 7) is $(\frac{4}{7},1)\subseteq [0,1]$, so $|(4,7)|\le |[0,1]|$.

Now Cantor-Bernstein does the rest.

It's a handy tool to not have to give an exact bijection between these two sets.

The injection $x\to x+5$ shows that we have an injection from [0, 1] into (4, 7), so by definition $|[0,1]\le |(4,7)|$.

Step 2

The injection $x\to \frac{x}{7}$ shows that we have an injection from (4, 7) into [0, 1] (the image of (4, 7) is $(\frac{4}{7},1)\subseteq [0,1]$, so $|(4,7)|\le |[0,1]|$.

Now Cantor-Bernstein does the rest.

It's a handy tool to not have to give an exact bijection between these two sets.

Andy Erickson

Answered 2022-05-25
Author has **3** answers

Step 1

The main strategy for these kinds of problems is that you want to show that one set is one-to-one and the other is one-to-one as well. You show this by finding a function the maps from one set to the other.

For this question let's say that the interval [0,1] is A, and the other (4,7) is B. Showing that $|A|\le |B|$ is saying that this function is one-to-one. To show this:

$[0,1]\mapsto (4,7)$, let's find a function that will map every input from [0, 1] to (4, 7). A function that would accomplish this would be $f(x)=x+5$. To show this you could make a simple diagram like: $0\mapsto 5\phantom{\rule{0ex}{0ex}}.5\mapsto 5.5\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\vdots \phantom{\rule{0ex}{0ex}}1\mapsto 6$

Step 2

For $(4,7)\mapsto [0,1]$, a potential function could be: $f(x)=\frac{x}{4}-1$. Again, drawing a diagram like:

$5\mapsto .25\phantom{\rule{0ex}{0ex}}\vdots $

By showing $[0,1]\mapsto (4,7)$ is one-to-one, then $|[0,1]|\le |(4,7)|$ and have shown $(4,7)\mapsto [0,1]$ meaning $|(4,7)|\le |[0,1]|$ is also one-to-one, then we can conclude using Schröder-Bernstein that $|[0,1]|=|(4,7)|$.

The main strategy for these kinds of problems is that you want to show that one set is one-to-one and the other is one-to-one as well. You show this by finding a function the maps from one set to the other.

For this question let's say that the interval [0,1] is A, and the other (4,7) is B. Showing that $|A|\le |B|$ is saying that this function is one-to-one. To show this:

$[0,1]\mapsto (4,7)$, let's find a function that will map every input from [0, 1] to (4, 7). A function that would accomplish this would be $f(x)=x+5$. To show this you could make a simple diagram like: $0\mapsto 5\phantom{\rule{0ex}{0ex}}.5\mapsto 5.5\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\vdots \phantom{\rule{0ex}{0ex}}1\mapsto 6$

Step 2

For $(4,7)\mapsto [0,1]$, a potential function could be: $f(x)=\frac{x}{4}-1$. Again, drawing a diagram like:

$5\mapsto .25\phantom{\rule{0ex}{0ex}}\vdots $

By showing $[0,1]\mapsto (4,7)$ is one-to-one, then $|[0,1]|\le |(4,7)|$ and have shown $(4,7)\mapsto [0,1]$ meaning $|(4,7)|\le |[0,1]|$ is also one-to-one, then we can conclude using Schröder-Bernstein that $|[0,1]|=|(4,7)|$.

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a) the set of sophomores taking discrete mathematics in your school

b) the set of sophomores at your school who are not taking discrete mathematics

c) the set of students at your school who either are sophomores or are taking discrete mathematics

Use these symbols:$\cap \cup$

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c) the set of students at your school who either are sophomores or are taking discrete mathematics

Use these symbols:

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I want to make sure that this is right with respect to the cardinality

Consider the following sets: $A=\{0,1\}$, $B=\{\{0,1\}\}$ and $C=A\cup B$. Enumerate the following sets and report their cardinality:

No. 1: ${2}^{C}$

No. 2: $C\times C$ (cross-product)

No. 1: $\{\},\{0\},\{0,1\},\{1\},\{0,0\},\{0,1\},\{1,0\},\{1,1\}$, so we have 8, so ${2}^{8}$ would be 256 and that would be the cardinality.

No. 2: $\{0\},\{0,1\},\{1\},\{0,0\},\{0,1\},\{1,0\},\{1,1\},\{0\},\{0,1\},\{1\},\{0,0\},\{0,1\},\{1,0\},\{1,1\}$ so it would be 14 for the cardinality.

I just want to make sure if these are right, and if not, find out where I went wrong.

Consider the following sets: $A=\{0,1\}$, $B=\{\{0,1\}\}$ and $C=A\cup B$. Enumerate the following sets and report their cardinality:

No. 1: ${2}^{C}$

No. 2: $C\times C$ (cross-product)

No. 1: $\{\},\{0\},\{0,1\},\{1\},\{0,0\},\{0,1\},\{1,0\},\{1,1\}$, so we have 8, so ${2}^{8}$ would be 256 and that would be the cardinality.

No. 2: $\{0\},\{0,1\},\{1\},\{0,0\},\{0,1\},\{1,0\},\{1,1\},\{0\},\{0,1\},\{1\},\{0,0\},\{0,1\},\{1,0\},\{1,1\}$ so it would be 14 for the cardinality.

I just want to make sure if these are right, and if not, find out where I went wrong.

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I have this formula for the density function:

$f(x)=\frac{{\mu}^{x}}{x!}\cdot {\mathrm{exp}}^{-\mu}$

$\mu =2$ because 2 days pass between fires I think but I am not sure what X should be.

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I have this formula for the density function:

$f(x)=\frac{{\mu}^{x}}{x!}\cdot {\mathrm{exp}}^{-\mu}$

$\mu =2$ because 2 days pass between fires I think but I am not sure what X should be.

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