# Open set in topology

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.

• What is your definition of an open set in $\mathbb{R}?$ May 25 '20 at 9:30
• 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). May 25 '20 at 9:30
• @Quimey. Agree. May 25 '20 at 9:36
• @SahibaArora For any $\varepsilon>0$ there exists $x_0$ in $(a,b)\subset \Bbb R$ such that $x_0-a < \varepsilon$ and $b-x_0 < \varepsilon$. May 25 '20 at 10:06

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.