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Let $\mathcal{B}$ be base for a topology $\mathcal{T}$.

From two following definitions of base and neighborhood respectively, does it imply that $\mathcal{B}$ is uncontable?
Is neighborhood really similar to open set in $\Bbb R$?
Here base is said to contain each neighborhood of $x$, since for each neighborhood it is possible to choose a set from $\mathcal{B}$.

$\mathcal{B}$ is a base for topology $\mathcal{T}$ iff $\mathcal{B}$ is a subfamily of $\mathcal{T}$ and for each point $x$ of the space, and each neighborhood $U$ of x, there is a member $V$ of $\mathcal{B}$ such that $x\in V\subset U$.

and

A set $U$ in a topological space is a neighborhood of a point $x$ iff $U$ contains an open set to which $x$ belongs.

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  • $\begingroup$ What is your definition of an open set in $\mathbb{R}?$ $\endgroup$ May 25 '20 at 9:30
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    $\begingroup$ There is a topology with only two open sets (empty and all space), There is only one neighborhood (the whole space) and there is a base with only one element (the whole space). $\endgroup$
    – Quimey
    May 25 '20 at 9:30
  • $\begingroup$ @Quimey. Agree. $\endgroup$
    – flowian
    May 25 '20 at 9:36
  • $\begingroup$ @SahibaArora For any $\varepsilon>0$ there exists $x_0$ in $(a,b)\subset \Bbb R$ such that $x_0-a < \varepsilon$ and $b-x_0 < \varepsilon$. $\endgroup$
    – flowian
    May 25 '20 at 10:06
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From two following definitions of base and neighborhood respectively, does it imply that $\mathcal{B}$ is uncontable?

No, even for the usual topology on $\mathbb{R}$ . For $x \in \mathbb{R}$ and $\epsilon > 0$, write $B(x,\epsilon) := ]x - \epsilon, x + \epsilon[$ for the open ball centered at $x$ vith radius $\epsilon$. Now consider $$\mathcal{B} := \big\{B(q,\frac{1}{n}) \ \big| q \in \mathbb{Q}, \ n\in \mathbb{N} \big\}$$

$\mathcal{B}$ is a basis for the topology, and is countable.

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