Exercise 17.14

Question

In the finite complement topology on ${\mathbb{R}}$, to what point or points does the sequence $x_n = 1/n$ converge?

Dan Whitman · Accepted Answer

We claim that this sequence converges to every point in ${\mathbb{R}}$.
\begin{proof}
  Suppose that this is not the case so that there is point $a \in {\mathbb{R}}$ where the sequence does not converge to $A$.
  Then there is an open set $U$ containing $a$ such that, for every $N \in {{\mathbb{Z}}_+}$, there is an $n \geq N$ where $x_n 
otin U$.
  It is easy to see that $x_n 
otin U$ for an infinite number of $n \in {{\mathbb{Z}}_+}$.
  For, if this were not the case, then there would be an $N \in {{\mathbb{Z}}_+}$ where $x_n \in U$ for every $n \geq N$.
  We know, though, that there must be an $n \geq N$ where $x_n 
otin U$.

Moreover, clearly, every $x_n$ in the sequence is distinct so that there are an infinite number of points not in $U$.
  Since each of these points are still in $X$, we have that $X - U$ is infinite.
  As this is the finite complement topology and $U$ is open, this can only be the case if $X - U = X$ itself, in which case it would have been that $U = \varnothing$ since $U \subset X$.
  This is not possible since $U$ contains $a$.
  So it seems that a contradiction has been reached, which shows the desired result.
\end{proof}

In fact, this is true for any sequence for which the image of the sequence  $\left\{x_n \mid n \in {{\mathbb{Z}}_+}ight\}$ is infinite.
This is to say that any such sequence converges to every point of ${\mathbb{R}}$.
Note also that this shows that the finite complement topology on ${\mathbb{R}}$ is not a Hausdorff space by the contrapositive of Theorem~17.10.

Exercise 17.14

Answers

Comments

Exercise 17.14

Answers

Comments

Add answer