Negation of statement involving cardinalities

[psie]

TL;DR Summary

I have a hard time understanding the negation of a statement involving cardinalities in Folland's real analysis book.

In Folland's real analysis book, he defines the following expressions: $$\operatorname{card}(X)\leq\operatorname{card}(Y),\quad \operatorname{card}(X)=\operatorname{card}(Y),\quad \operatorname{card}(X)\geq\operatorname{card}(Y),$$to mean there exists an injection, bijection or surjection from $X$ to $Y$ respectively. Then he defines $$\operatorname{card}(X)<\operatorname{card}(Y),\quad \operatorname{card}(X)>\operatorname{card}(Y),$$to mean there exists an injection but no bijection or a surjection but no bijection from $X$ to $Y$ respectively. Later in the section, he states the following corollary:

Corollary 0.13: If $\operatorname{card}(X)\geq\mathfrak{c}$, then $X$ is uncountable.

He says the converse of this statement is the so-called continuum hypothesis. The converse would be, taking the contrapositive,

If $\operatorname{card}(X)\not\geq\mathfrak{c}$, then $X$ is countable.

My question is; is $\operatorname{card}(X)\not\geq\mathfrak{c}\iff \operatorname{card}(X)<\mathfrak{c}$? How can one show this equivalence?

 

[jedishrfu]

Since cardinalities are whole numbers ie natural numbers including zero then the $>$ and $<$ and $=$ relationships and their definitions apply.

so the expression card(X) $>=$ card(Y) means card(X) $>$ card(Y) or card(X) $=$ card(Y) and its inverse would be card(X) $<$ card(Y).

 

[FactChecker]

Since cardinalities are whole numbers ie natural numbers including zero then the $>$ and $<$ and $=$ relationships and their definitions apply.

That makes the OP statement "he defines the following expressions $\textrm{card}(X) \le \textrm{card}(Y)$"a little confusing.
It seems like either this "definition" is wrong or the function card() does not have numerical values at this point.

 

[psie]

Since cardinalities are whole numbers ie natural numbers including zero then the $>$ and $<$ and $=$ relationships and their definitions apply.

Thanks. Probably one could say cardinalities are ordinals. Anyway, Folland does not define cardinalities as ordinals. He gives no definition at all for the object itself. He does define cardinalities for finite sets as being the number of elements in that set, in which case your reasoning applies. The only thing given are the definitions in terms of $\leq, =,\geq, <$ and $>$. I wonder if it’s possible to deduce the equivalence using those.

 

[jedishrfu]

That makes the OP statement "he defines the following expressions $\textrm{card}(X) \le \textrm{card}(Y)$"a little confusing.
It seems like either this "definition" is wrong or the function card() does not have numerical values at this point.

I assumed he was using it in the context of set theory as the number of elements in a set.

 

[FactChecker]

I assumed he was using it in the context of set theory as the number of elements in a set.

Sounds reasonable. I am not sure what is being "defined" in the OP -- is it the function card() or the inequality symbols? Or both? There are problems with any of those options. Maybe the OP uses the word "defined" loosely.

 

[jedishrfu]

I thought it was the equality symbols, and he wanted to know how they applied to the card() function.

Perhaps @fresh_42 could comment here on the OPs questions.

 

[psie]

Here's some food for thought.

If $\mathrm{card}(X)\geq\mathfrak{c}$, then by the partition principle (i.e. there's an injection $X\to Y$ iff there's a surjection from $Y\to X$; this requires axiom of choice by the way), $\mathfrak{c}\leq \mathrm{card}(X)$. Now, if there is no injection between the set with cardinality $\mathfrak{c}$ to $X$, i.e. if not $\mathfrak{c}\leq \mathrm{card}(X)$, then by total ordering (i.e. either $\mathfrak{c}\leq \mathrm{card}(X)$ or $\mathfrak{c}\geq \mathrm{card}(X)$), we have $\mathfrak{c}\geq \mathrm{card}(X)$. And again by partition principle, $\mathrm{card}(X)\leq\mathfrak{c}$. I'm just not sure how to conclude $\mathrm{card}(X)<\mathfrak{c}$, i.e. the "strict inequality".

EDIT: I guess, if not $\mathfrak{c}\leq \mathrm{card}(X)$, then we must have $\mathfrak{c}>\mathrm{card}(X)$, i.e. an injection but no bijection. Because if a bijection would exist, then $\mathfrak{c}\leq\mathrm{card}(X)$ would be true after all. What I've learnt about this is that the continuum hypothesis has a certain version assuming the axiom of choice, since negating $\mathrm{card}(X)\geq\mathfrak{c}$ seems to depend on axiom of choice.

 

[Gavran]

In my opinion the partition principle can be expressed in two ways:
$\left|\mathrm{X}\right| \ge \left|\mathrm{Y}\right| \Leftrightarrow \left|\mathrm{Y}\right| \le \left|\mathrm{X}\right|$ where a bijection is included
and
$\left|\mathrm{X}\right| \gt \left|\mathrm{Y}\right| \Leftrightarrow \left|\mathrm{Y}\right| \lt \left|\mathrm{X}\right|$ where a bijection is excluded.

The negation of $\left|\mathrm{X}\right| \ge \left|\mathrm{Y}\right|$ will exclude a surjection and a bijection from $\mathrm{X} \to \mathrm{Y}$ what means that $\mathrm{X}$ only injects into $\mathrm{Y}$. Now by the partition principle there will be $\mathrm{X}$ only injects into $\mathrm{Y}$ if and only if $\mathrm{Y}$ only surjects onto $\mathrm{X}$.

 

[Hornbein]

He says the converse of this statement is the so-called continuum hypothesis. The converse would be, taking the contrapositive,

The contrapositive of a statement always has the same truth value as that statement. It cannot be the converse.

The contrapositive of if A then B is if not B then not A.

My hat it has three corners
Three corners has my hat
And if it has not three corners
Then it is not my hat.

The continuum hypothesis is "There is no set whose cardinality is strictly between that of the integers and the real numbers." There does not exist S such that Card(N) < Card(S) < Card(R).

The negation or converse is "There is at least one set whose cardinality is strictly between that of the integers and the real numbers." There exists S such that Card(N) < Card(S) < Card(R).

 

[pbuk]

He says the converse of this statement is the so-called continuum hypothesis. The converse would be, taking the contrapositive,

There seems to be some confusion here: the contrapositive is not the same as the converse.

The converse of

"if $\operatorname{card}(X)\geq\mathfrak{c}$, then $X$ is uncountable"​

is

"if $X$ is uncountable, then $\operatorname{card}(X)\geq\mathfrak{c}$".​

This is indeeed equivalent to the continuum hypothesis, as can be seen more clearly by rephrasing as "all uncountable sets have cardinalty at least that of the continuum".