# Inner Limit in Hausdorff Space by Open Neighborhoods

## Theorem

Let $\left \langle{C_n}\right \rangle_{n \in \N}$ be a sequence of sets in a Hausdorff topological space $\left({\mathcal X, \tau}\right)$.

Let $x \in \mathcal X$.

Let $\mho \left({x}\right) := \left\{ {V \in \tau:\ x \in V}\right\}$ denote the set of open neighborhoods of $x$.

Let $\mathcal N_\infty$ denote the set of cofinite subsets of $\N$:

$\mathcal N_\infty := \left\{ {N \subset \N: \N \setminus N \text{ is finite} }\right\}$

Then the inner limit of $\left \langle{C_n}\right \rangle_{n \in \N}$ is:

$\displaystyle \liminf_n C_n = \left\{ {x \in \mathcal X: \forall V \in \mho \left({x}\right): \exists N \in \mathcal N_\infty: \forall n \in N: C_n \cap V \ne \varnothing}\right\}$

or equivalently:

$\displaystyle \liminf_n C_n = \left\{{x \in \mathcal X: \forall V \in \mho \left({x}\right): \exists N_0 \in \N: \forall n \ge N_0: C_n \cap V \ne \varnothing}\right\}$

## Proof

If $x \in \liminf_n C_n$ then there exist a sequence $\left \langle{x_k}\right \rangle_{n \in \N}$ such that $x_k \to x$ while:

$x_k \in C_{n_k}$

and

$\left \langle{n_k}\right \rangle_{k \in \N} \subseteq \N$ is a strictly increasing sequence of indices.

For any $V \in \mho \left({x}\right)$ there exists $N_0 \in\N$ such that for all $i \ge N_0$:

$x_i \in V$

and:

$x_i \in C_{n_i}$

Thus:

$C_{n_i} \cap V \ne \varnothing$

Therefore $x \in \left\{{x \in \mathcal X: \forall V \in \mho \left({x}\right): \exists N_0 \in \N: \forall n \ge N_0: C_n \cap V \ne \varnothing}\right\}$.

$\Box$

Let $x \in \left\{ {x \in \mathcal X: \forall V \in \mho \left({x}\right): \exists N \in \mathcal N_\infty: \forall n \in N: C_n \cap V \ne \varnothing}\right\}$.

Thus:

$\forall V \in \mho \left({x}\right): \exists N \in \mathcal N_\infty: \forall n \in N: C_n \cap V \ne \varnothing$

Then there exists a strictly increasing sequence:

$\left \langle{n_k}\right \rangle_{k \in \N} \subseteq \N$

such that for every $V \in \mho \left({x}\right)$:

$\exists x_k \in C_{n_k}\cap V$.

Hence $x_k \to x$ in the topology $\tau$.

$\blacksquare$