Which of the following statement is True ?
$a)$ The family of subsets $\Big\{ A_n=(\frac{-1}{n},\frac{1}{n}): n=1,2….\Big\}$ satisfies the finite intersection property
$b)$ If $f:\mathbb{R} \rightarrow X$ is continuous ,where $\mathbb{R}$ is the given the usual topology and $(X,\tau)$ is a Hausdorff $(T_1)$ space ,then $f$ is an one-one function
My attempt : option a) is false because $(\frac{-1}{2},\frac{1}{2}) \cap (\frac{-1}{3},\frac{1}{3}) = ∅$
option b) is true take $f(x) = x$
Is my logic is correct or not ?
any hints/solution wil bee appreciated
thanks in advance
Best Answer
The statement $a)$ is true. Indeed, let's take a finite number $\left(-\frac{1}{n_1}, \frac{1}{n_1}\right)$, ..., $\left(-\frac{1}{n_k}, \frac{1}{n_k}\right)$ of such intervals. Then for $N = \max{n_i}$, you have that for every $i = 1, ..., k$, $$\left(-\frac{1}{N}, \frac{1}{N}\right) \subset \left(-\frac{1}{n_i}, \frac{1}{n_i}\right)$$
and then $$\left(-\frac{1}{N}, \frac{1}{N}\right) \subset \bigcap_{i=1}^k\left(-\frac{1}{n_i}, \frac{1}{n_i}\right)$$ so this intersection is non empty.
The statement $b)$, however is false. Indeed, $f : \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x)=0$ for all $x \in \mathbb{R}$ is continuous between two Hausdorff spaces but is not one-one.