Neighborhood Basis in Real Number Line is Infinite

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\R$ be the real number line considered as a metric space under the usual metric.

Let $a \in \R$ be a point in $\R$.

Let $\mathcal B_a$ be a basis for the neighborhood system of $a$.


Then $\mathcal B_a$ is an infinite set.


Proof

Suppose $\mathcal B_a$ be finite.

Let the elements of $\mathcal B_a$ be enumerated as $N_1, N_2, \ldots, N_n$.

For each $N_k \in \mathcal B_a$, let $B_{\epsilon_k} \left({a}\right)$ be the open $\epsilon_k$-ball of $a$ for some $\epsilon_k \in \R_{>0}$.

Let $\alpha = \min \left\{ {\epsilon_k: k \in \left\{ {1, 2, \ldots, n}\right\} }\right\}$.

Consider the open interval:

$I = \left({a - \beta \,.\,.\, a + \beta}\right)$

where $\beta = \dfrac \alpha 2$

By the method of construction of $I$, it follows that $B_{\epsilon_k} \left({a}\right) \nsubseteq I$ for any $k$.

Thus $I$ is a neighborhood of $a$ which does not have an element of $\mathcal B_a$ as a subset.

So $\mathcal B_a$ cannot be a basis for the neighborhood system of $a$.

Hence the result.

$\blacksquare$


Sources