Proof verification and understanding

Question

I am using Topology Without Tears by Sydney Morris

I am self studying it.

Theorem

If is an infinite set of cardinality ℵ prove that there are at least ℵ distinct topologies on it.

My thoughts on it.(Edited)

Let X be infinite.

P(X)=$2^{\aleph}$

Let A$\in P(X)$ .Define $T_A$={$\emptyset$,A,X}

$T_A$ is a topology.

Clearly $A\ne B \Rightarrow T_A\ne T_B$

So C $\in$ P(X), $T_C$= {$\emptyset$,C,X} is a topology

By A1.2.1 A~C So U={$T_a$ :A$\in $P(X)}is uncountable

And for our bijection

f:X$\mapsto $ P(X)

If A$\ne C \Rightarrow T_A\ne T_C$ Then B$\in $P(X) and $T_B$={$\emptyset$,B,X} is uncountable.

By A1.2.1 A~B and U={$T_A$ :A$\in $P(X)}is uncountable

Any help to make me understand this stuff a bit easier would be helpful

Left out first theorem on pg 39 cause answers found. I am wondering if there are other ways to do it inductively.

Answer 1

You’ve actually constructed $2^\aleph$ distinct topologies on $X$, which is what you actually need: you slightly misquoted the problem. I will note, however, that many of them are homeomorphic to one another: $\langle X,T_A\rangle$ is homeomorphic to $\langle X,T_B\rangle$ if and only if $|A|=|B|$. If $\aleph=\aleph_1$, for instance, you actually have only $\aleph_0$ pairwise non-homeomorphic topologies, far fewer than $2^{\aleph_1}$. (I’ve used $T_A$ for your $T_a$, since there is no necessary connection between $a$ and $A$: they are distinct symbols, and you want the name of the topology to refer to the subset of $X$ used to define it.) However, it appears from his hint that you’ve done what Morris had in mind.

Nevertheless, some of what you’ve said is incorrect.

• $T_A$ is not infinite: it has either $2$ or $3$ elements, depending on whether $A\in\{\varnothing,X\}$ or not.
• $U$ is indeed uncountable, since $\aleph$ is infinite, but that isn’t what you need to prove: you need to show that $|U|\ge|X|$.
• No matter what the set $A$ is, there is no bijection from $A$ to $\wp(A)$: it is a general theorem that $|A|<|\wp(A)|$ for all sets $A$. It is also not clear what any such function would have to do with the problem. The relevant bijection is from $\wp(X)$ to $U$ and is given by the map $\varphi(A)=T_A$.

Added: You’re really almost done once you define the topologies $T_A$. You could write it up like this, for instance:

For each $A\in\wp(X)\setminus\{\varnothing,X\}$ let $T_A=\{\varnothing,A,X\}$; clearly $T_A$ is a topology on $X$ with $3$ open sets. Moreover, if $A,B\in\wp(X)\setminus\{\varnothing,X\}$, and $A\ne B$, then $T_A\ne T_B$, since $A\in T_A\setminus T_B$. Thus, if $U=\big\{T_A:A\in\wp(X)\setminus\{\varnothing,X\}\big\}$, the map $\varphi:\wp(X)\setminus\{\varnothing,X\}\to U:A\mapsto T_A$ is a bijection. Thus, $|U|=|\wp(X)\setminus\{\varnothing,X\}|\overset{(*)}=|\wp(X)|=2^{|X|}=2^\aleph$, where $(*)$ holds because $\wp(X)$ is infinite.

Note that $T_\varnothing=T_X$, so you can’t keep both of them in $U$ if you want $\varphi$ to be injective; I chose simply to delete both of them. You could keep one of them, but then the proof that $T_A\ne T_B$ when $A\ne B$ has to be broken into two cases, since the topology $T_\varnothing$ (or $T_X$) has to be handled separately.