The weight is a monotonic topological property

Question

Achtung

If $X$ is a topological space for convenience we define $$ \omega(X):=\min\{|\mathcal{B}|:\mathcal{B}\quad\text{is a basis for }X\} $$ that is well define since the cardinal are well ordered.

Definition (1)

The weight $w(X)$ of a topological spaces $X$ is the following quantity $$ w(X):=\omega(X)+\aleph_0 $$ that is well defined, since $\omega(X)$ is defined.

Statement (2)

If $X$ and $Y$ are two homeomorphic topological spaces then $w(X)=w(Y)$.

Proof. So let be $X$ and $Y$ two topological spaces and we suppose that $h:X\rightarrow Y$ is an homeomorphism thus if $\mathcal{B}$ and $\mathcal{D}$ are bases of $X$ and $Y$ then the collection $\mathcal{B}'=\{h(B):B\in\mathcal{B}\}$ and $\mathcal{D}'=\{h^{-1}(D):D\in\mathcal{D}\}$ are bases of $Y$ and $X$ and so $\omega(Y)\le|\mathcal{B}'|$ and $\omega(X)\le|\mathcal{D}'|$. So since the $h$ is bjective it results that $|\mathcal{B}'|=|\mathcal{B}|$ and $|\mathcal{D}'|=|\mathcal{D}|$, thus if we choose $\mathcal{B}$ and $\mathcal{D}$ such that $|\mathcal{B}|=\omega(X)$ and $|\mathcal{D}|=\omega(Y)$ then it results that $\omega(X)=\omega(Y)$ and so $w(X):=\omega(X)+\aleph_0=\omega(Y)+\aleph_0=:w(Y)$.

Statement (3)

If $X$ is a topological space and $Y\subseteq X$ then $w(Y)\le w(X)$.

Proof. So let be $X$ a topological space and $Y$ a subspace. So if $\mathcal{B}$ is a base for $X$ then the collection $\mathcal{D}=\{B\cap Y:B\in\mathcal{B}\}$ is a base for $Y$ and so if for any $D\in\mathcal{D}$ we pick $B_D\in\mathcal{B}$ such that $D=B_D\cap Y$ then it is clear that the relation $\phi$ between $\mathcal{D}$ and $\mathcal{B}$ defined as $D\phi B\Leftrightarrow B=B_D$ is a injective function and so $|\mathcal{D}|\le|\mathcal{B}|$; thus if we pick $\mathcal{B}$ such that $|\mathcal{B}|=\omega(X)$ then $\omega(Y)\le\omega(X)$, from which it is clear that $w(Y):=\omega(Y)+\aleph_0\le\omega(X)+\aleph_0=:w(X)$.

So are the proofs correct? if not how to prove the two statements? could someone help me, please?

Answer 1

Your proofs are correct, but they could be stated a little more clearly, especially the second one. I might write them up like this, for instance. (Note, though, that to some extent this is just a matter of taste, and what I find clearer or neater might seem less so to you.)

For the first result let $h:X\to Y$ be a homeomorphism, let $\mathscr{B}$ be a base for $X$, and let $\mathscr{B}'=\{h[B]:B\in\mathscr{B}\}$; it is straightforward to verify that $\mathscr{B}'$ is a base for $Y$. Since $h$ is a homeomorphism, the function $\mathscr{B}\to\mathscr{B}':B\mapsto h[B]$ is a bijection, $|\mathscr{B}'|=|\mathscr{B}|$, and it follows that $\omega(Y)\le\omega(X)$. But $h^{-1}:Y\to X$ is also a homeomorphism, so we have similarly that $\omega(X)\le\omega(Y)$, and we conclude that $\omega(X)=\omega(Y)$ and hence that $w(X)=w(Y)$.

For the second result let $\mathscr{B}$ be a base for $X$, and let $\mathscr{B}'=\{B\cap Y:B\in\mathscr{B}\}$; $\mathscr{B}'$ is a base for $Y$. For each $D\in\mathscr{B}'$ fix a $B_D\in\mathscr{B}$ such that $D=B\cap Y$. Then the function $\mathscr{B}'\to\mathscr{B}:D\mapsto B_D$ is injective, so $|\mathscr{B}'|\le|\mathscr{B}|$, and it follows that $\omega(Y)\le\omega(X)$ and hence that $w(Y)\le w(X)$.