Let $B:=\{B_{\epsilon}(x):\forall x\in \Bbb{R}^3,\epsilon>1\}$. Does $B$ form a basis for a topology?

Question

Let $X=\mathbb{R}^3$, and let $B$ be the family of open $\epsilon$-balls {$B_{\epsilon}(x):x\in X,\epsilon>1$}.Does $B$ form a basis for a topology? (describe generated topology) can someone pls help me with this problem? Im completely lost....

Answer 1

No, it doesn't form the basis of a topology. Take several balls $B_\varepsilon(x)$ such that their intersection $I$ is small; more precisely, such the diameter if $I$ is smaller than $2$. Then $I$ would be an open set, but $I$ contains no ball whose radius is greater than $1$.

Answer 2

(I). A necessary and sufficient condition for a family $B$ of subsets of a set $X$ to be a base (basis) for a topology on $X$ is:

(I-i). $\;\cup B=X.\;$ (I.e. every $p\in X$ belongs to at least one member of $B$), and

(I-ii). $\; \forall f_1,f_2\in B\; \forall p\in f_1\cap f_2\;( \exists f\in B\;(p\in f\subset f_1\cap f_2)).$

(II). Take $r>0$ such that $2r+r^2<1.$ Let $y_1=(-1,0,0)$ and $y_2=(1,0,0).$ Let $f_1=B(y_1,1+r)$ and $f_2=B(y_2,1+r).$ Let $p=(0,0,0).$

We have (obviously) $f_1,f_2\in B$ and $p\in f_1\cap f_2.$

(II-i). For any $f=B(x,\varepsilon)\in B$ with $\varepsilon >1$ and $x=(x_1,x_2,x_3), $ both $x^+= (x_1 + 1,x_2,x_3)$ and $x^-=(x_1-1,x_2,x_3)$ belong to $f$ and their distance apart, $d(x^+,x^-),$ is $2.$

(II-ii). If $p\in f\in B$ then there exists $q\in f$ such that $d(p,q)\geq 1.$ Proof: By (II-i) take $x^+, x^- \in f$ with $d(x^+,x^-)=2.$ Then $d(x^+,p)+d(p,x^-)\geq d(x^+,x^-)=2$. So there exists $q\in \{x^+,x^-\}\subset f$ such that $d(p,q)\geq 1.$

(II-iii). Now suppose $f\in B$ and $p\in f\subset f_1\cap f_2.$ By (II-ii) take $q \in f$ such that $d(p,q)\geq 1.$ Let $q=(q_1,q_2,q_3).$ $$ \text { Now } q\in f_1. \text { So } (1+r)^2> (d(q,y_1))^2=(q_1+1)^2+q_2^2+q_3^2.$$ $$\text { And } q\in f_2. \text { So } (1+r)^2>(d(q,y_2))^2= (q_1-1)^2 +q_2^2+q_3^2.$$ Adding the above two inequalities, we have $$2(1+r)^2>2(q_1^2+q_2^2+q_3^2)+2$$ which implies $$2r+r^2>q_1^2+q_2^2+q_3^2=(d(q,p))^2.$$ But this contradicts $2r+r^2<1\leq 1\leq (d(p,q))^2.$

Therefore there does not exist $f\in B$ such that $p\in f\subset f_1\cap f_2.$ So condition (I-ii) is not met and $B$ is not a base (basis) for a topology on $\Bbb R^3.$