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I have just started learning metric space and our professor has defined two terminologies:

$1.$Base of a topological space $(X,\tau)$ is defined to be a set $\scr B\subset \tau$ such that for each $U\in \tau$ and $x\in U$, $\exists B\in \scr B$ such that $x \in B \subset U$.

$2.$Local base at a point $x\in X$ in a topological space is a collection of neighbourhoods $(v_\alpha)_{\alpha \in \lambda} $of $x$ such that for any nbd $U_x$ of $x$,some $v_\alpha \subset U_x$.

Now I have some question(follow these definitions) regarding what are the connections between local base and a global base?I need some properties that would help me to understand the connection between these two things and also enable me to work freely with these.Can somebody please help me with them? Since I have yet not learnt topology,I do not know much of it.

Paul Frost
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Kishalay Sarkar
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    It might help to consider the standard example for $\mathbb{R}^n$. A global base would be the set ${B_{\varepsilon} (x) \vert x \in \mathbb{R}^n, \varepsilon > 0}$ i.e. the set of open balls, whereas for any $x \in \mathbb{R}^n$ a choice of local base would be ${B_{\varepsilon} (x) \vert \varepsilon > 0}$ i.e. the set of open balls centered around $x$. – G. Chiusole Feb 13 '20 at 14:13
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    Also, formulating a concrete question might make it easier to give a satisfactory answer. – G. Chiusole Feb 13 '20 at 14:16
  • @G.Chiusole Actually I want to get a summary of their basic properties(I will prove them myself but atleast I must have something to start with,the problem with books is they define little differently).Anyways the list I am talking about may contain properties like second countable implies first countable,union of local bases is a global base,basic open sets containing x form a local basis at x etc. – Kishalay Sarkar Feb 13 '20 at 14:39
  • My answer to Confusion Regarding Munkres's Definition of Basis for a Topology might be of use. – Dave L. Renfro Feb 13 '20 at 15:05
  • In 2., are the neighborhoods required to be open? – Paul Frost Feb 13 '20 at 15:12

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Let be $(X,\mathcal{T})$ a topological space: first we observe that the collection $$ \mathfrak{B}=\{\mathcal{B}\subseteq\mathcal{T}:\text{ basis for }\mathcal{T}\} $$ of the basis $\mathcal{B}$ for $\mathcal{T}$ is not empty because trivially $\mathcal{T}\in\mathfrak{B}$; analogously for any point $x$ of $X$ the collection $$ \mathcal{V}(x)=\{V_x\subseteq X: V_x\text{ is a neighborhood of } x\} $$ of the neighborhood of $x$ is not empty because trivially $X\in\mathcal{V}(x)$ so the collecion $$ \mathfrak{V}(x)=\{\mathcal{B}(x)\subseteq\mathcal{P}(X):\mathcal{B}(x)\quad\text{is a local basis for } x\} $$ is not empty because trivially $\mathcal{V}(x)\in\mathfrak{V}(x)$.

Well it is not complicated to prove that for any $\mathcal{B}\in\mathfrak{B}$ and for any $x\in X$ the collection $$\mathcal{B}(x)=\{B\in\mathcal{B}:x\in B\} $$ is a local basis for $x$; analogously it is not complicated to prove that for any $x\in X$ the collection $$ \bigcup_{x\in X}\mathfrak{A}(x) $$

is a basis for $\mathcal T$ when $\mathfrak A(x)$ is the set $$ \mathfrak A(x):=\{\mathcal A(x)\in\mathfrak V(x):\mathcal A(x)\text{ is an open local base for }x\} $$ for all $x\in X$.

Antonio Maria Di Mauro
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