Difference between revisions of "Notes:Basis for a topology/Attempt 2"

Latest revision as of 10:14, 8 August 2016

Overview

Last time I tried to merge the definitions, which got very confusing! This time I shall treat them as separate things and go from there.

Definitions

Here we will use GBasis for a generated topology (by a basis) and TBasis for a basis of an existing topology.

GBasis

Let $X$ be a set and $\mathcal{B}\subseteq\mathcal{P}(X)$ be a collection of subsets of $X$. Then we say:

• $\mathcal{B}$ is a GBasis if it satisfies the following 2 conditions:
  1. $\forall x\in X\exists B\in\mathcal{B}[x\in B]$ - every element of $X$ is contained in some GBasis set.
  2. $\forall B_1,B_2\in\mathcal{B}\forall x\ B_1\cap B_2\exists B_3\in\mathcal{B}[B_1\cap B_2\ne\emptyset\implies(x\in B_3\subseteq B_1\cap B_2)]$^{[Note 1][Note 2]}

Then $\mathcal{B}$ induces a topology on $X$.

Let $\mathcal{J}_\text{Induced}$ denote this topology, then:

• $\forall U\in\mathcal{P}(X)\big[U\in\mathcal{J}_\text{Induced}\iff\big(\forall p\in U\exists B\in\mathcal{B}[p\in B\wedge B\subseteq U]\big)\big]$

TBasis

Suppose $(X,\mathcal{ J })$ is a topological space and $\mathcal{B}\subseteq\mathcal{P}(X)$ is some collection of subsets of $X$. We say:

• $\mathcal{B}$ is a TBasis if it satisfies both of the following:
  1. $\forall B\in\mathcal{B}[B\in\mathcal{J}]$ - all the basis elements are themselves open.
  2. $\forall U\in\mathcal{J}\exists\{B_\alpha\}_{\alpha\in I}[\bigcup_{\alpha\in I}B_\alpha=U]$

If we have a TBasis for a topological space then we may talk about its open sets differently:

• $\forall U\in\mathcal{P}(X)\big[U\in\mathcal{J}\iff\big(\forall p\in U\exists B\in\mathcal{B}[p\in B\wedge B\subseteq U]\big)\big]$

Facts

1. $\mathcal{B}$ is a GBasis of $X$ inducing $(X,\mathcal{J}')$ $\implies$ $\mathcal{B}$ is a TBasis for the $(X,\mathcal{J}')$
2. $(X,\mathcal{ J })$ is a topological space with a TBasis $\mathcal{B}$ $\implies$ $\mathcal{B}$ is a GBasis

Result

1. Lemma: $\mathcal{B}$ is a TBasis of $(X,\mathcal{J})$ $\implies$ $\mathcal{B}$ is a GBasis (inducing $(X,\mathcal{J}')$) $\implies$ $\mathcal{B}$ is a TBasis of $(X,\mathcal{J}')$
2. Proposition: $\mathcal{J}=\mathcal{J'}$ (using the method of proof of fact 1 part 3 (the bit with the warning - as it isn't needed there))
3. Theorem: $\mathcal{B}$ is a TBasis of $(X,\mathcal{J})$ $\implies$ $\mathcal{B}$ is a GBasis (inducing $(X,\mathcal{J})$) $\implies$ $\mathcal{B}$ is a TBasis of $(X,\mathcal{J})$
4. Proposition: $\mathcal{B}$ is a Tbasis of $(X,\mathcal{J})$ $\iff$ $\mathcal{B}$ is a GBasis inducing $(X,\mathcal{J})$

Proof of facts

1. $\mathcal{B}$ is a GBasis of $X$ inducing $(X,\mathcal{J}')$ $\implies$ $\mathcal{B}$ is a TBasis for the $(X,\mathcal{J}')$
   • Let $\mathcal{B}$ be a GBasis, suppose it generates the topological space $(X,\mathcal{J}')$, we will show it's also a TBasis of $(X,\mathcal{J}')$
     1. $\forall B\in\mathcal{B}[B\in\mathcal{J}']$ must be shown
        • Let $B\in\mathcal{B}$ be given.
          • Recall $U\in\mathcal{J}'\iff\forall x\in U\exists B'\in\mathcal{B}[x\in B'\wedge B'\subseteq U]$
          • Let $x\in B$ be given.
            • Choose $B':=B$
              • $x\in B'$ by definition, as $x\in B'=B$ and
              • we have $B\subseteq B$, by the implies-subset relation, if and only if $\forall p\in B[p\in B]$ which is trivial.
          • Thus $B\in\mathcal{J}'$
        • Since $B\in\mathcal{B}$ was arbitrary we have shown $\forall B\in\mathcal{B}[B\in\mathcal{J}']$
     2. $\forall U\in\mathcal{J}'\exists\{B_\alpha\}_{\alpha\in I}[\bigcup_{\alpha\in I}B_\alpha=U]$ must be shown
        • Let $U\in\mathcal{J}'$ be given.
          • Then, as $\mathcal{B}$ is a GBasis, by definition of the open sets generated: $\forall x\in X\exists B\in\mathcal{B}[x\in B\wedge B\subseteq U]$
          • Let $p\in U$ be given
            • Define $B_p$ to be the $B$ that exists such that $p\in B_p$ and $B_p\subseteq U$
          • We now have an .... thing, like a sequence but arbitrary, $(B_p)_{p\in U}$ or $\{B_p\}_{p\in U}$ - I need to come down on a notation for this - such that:
            • $\forall B_p\in(B_p)_{p\in U}[p\in B_p\wedge B_p\subseteq U]$
          • We must now show $\bigcup_{p\in U}B_p=U$, we can do this by showing $\bigcup_{p\in U}B_p\subseteq U$ and $\bigcup_{p\in U}B_p\supseteq U$
            1. $\bigcup_{p\in U}B_p\subseteq U$
               • Using the union of subsets is a subset of the union we see:
                 • $\bigcup_{p\in U}B_p\subseteq U]$
            2. $U\subseteq\bigcup_{p\in U}B_p$
               • Using the implies-subset relation we see: $U\subseteq\bigcup_{p\in U}B_p\iff\forall x\in U[x\in\bigcup_{p\in U}B_p]$, we will show the RHS instead.
                 • Let $x\in U$ be given
                   • Recall, by definition of union, $x\in\bigcup_{p\in U}B_p\iff\exists q\in U[x\in B_q]$
                     • Choose $q:=x$ then we have $x\in B_x$ (as $p\in B_p$ is one of the defining conditions of choosing each $B_p$!)
               • Thus $U\subseteq\bigcup_{p\in U}B_p$
          • We have shown $\bigcup_{p\in U}B_p=U$
        • Since $U\in\mathcal{J}'$ was arbitrary we have shown this for all $U\in\mathcal{J}'$
     3. We must now show that $\mathcal{B}$ is a TBasis of $(X,\mathcal{J}')$ specifically. Warning:No we do not, but what I did here is useful in proving something else later
        • Recall that, by definition of $\mathcal{B}$ being a GBasis:
          • $\forall U\in\mathcal{P}(X)\big[U\in\mathcal{J}'\iff\big(\forall p\in U\exists B\in\mathcal{B}[p\in B\wedge B\subseteq U]\big)\big]$
        • Recall that, by definition of $\mathcal{B}$ being a TBasis we can talk about open sets as:
          • $\forall U\in\mathcal{P}(X)\big[U\in\mathcal{J}\iff\big(\forall p\in U\exists B\in\mathcal{B}[p\in B\wedge B\subseteq U]\big)\big]$ (for the topology it is a TBasis of, $\mathcal{J}$)
        • We wish to show that $\mathcal{J}'=\mathcal{J}$
          • Let $U\in\mathcal{P}(X)$ be given.
            • Then by definition of $\mathcal{B}$ being a GBasis:
              • $U\in\mathcal{J}'\iff\big(\forall p\in U\exists B\in\mathcal{B}[p\in B\wedge B\subseteq U]\big)$
            • But by definition of $\mathcal{B}$ being a TBasis (of $\mathcal{J}$):
              • $\big(\forall p\in U\exists B\in\mathcal{B}[p\in B\wedge B\subseteq U]\big)\iff U\in\mathcal{J}$
            • Combining these we see:
              • $U\in\mathcal{J}'\iff\big(\forall p\in U\exists B\in\mathcal{B}[p\in B\wedge B\subseteq U]\big)\iff U\in\mathcal{J}$
            • Thus: $U\in\mathcal{J}'\iff U\in\mathcal{J}$
          • We have shown $\forall U\in\mathcal{P}(X)[U\in\mathcal{J}'\iff U\in\mathcal{J}]$
          • For any topology on $X$, $\mathcal{K}$ we require: $\mathcal{K}\subseteq\mathcal{P}(X)$. So,
            • we know: $\forall V\in\mathcal{J}[V\in\mathcal{P}(X)]$ (from the implies-subset relation), and $\forall V'\in\mathcal{J}'[V'\in\mathcal{P}(X)]$
          • We want to show $\mathcal{J}=\mathcal{J}'$ remember, we will do this by showing $\mathcal{J}\subseteq\mathcal{J}'$ and $\mathcal{J}'\subseteq\mathcal{J}$
            1. Showing $\mathcal{J}\subseteq\mathcal{J}'$, using the implies-subset relation this is the same as showing $\forall V\in\mathcal{J}[V\in\mathcal{J}']$
               • Let $V\in\mathcal{J}$ be given.
                 • Then $V\in\mathcal{P}(X)$
                 • But we know: $\forall W\in\mathcal{P}(X)[W\in\mathcal{J}\iff W\in\mathcal{J}']$, so we can apply this using $V$.
                   • We see $V\in\mathcal{J}\iff V\in\mathcal{J}'$
                     • But by definition $V\in\mathcal{J}$ !
                   • So $V\in\mathcal{J}'$
               • Since $V\in\mathcal{J}$ was arbitrary we have shown that for all such $V$ that $V\in\mathcal{J}'$
            2. Showing $\mathcal{J}'\subseteq\mathcal{J}$ is almost exactly the same, so rather than copy and paste and change a few symbols, I'll just say "same as above".
        • We have shown that $\mathcal{J}=\mathcal{J}'$
   • This completes the proof.
   • We have shown that if $\mathcal{B}$ is a GBasis that not only is it a TBasis

Notes

1. Jump up ↑ Note that $x\in B_3\subseteq B_1\cap B_2$ is short for:
   • $x\in B_3\wedge B_3\subseteq B_1\cap B_2$
2. Jump up ↑ Note that if $B_1\cap B_2$ is empty (they do not intersect) then the logical implication is true regardless of the RHS of the $\implies$} sign, so we do not care if we have $x\in B_3\wedge B_3\subseteq B_1\cap B_2$! Pick any $x\in X$ and aany $B_3\in\mathcal{B}$!