Can convergence be seen as a form of continuity?

Question

I will abbreviate "it holds that" to "iht" and "such that" to "sth."

The following questions are motivated by curiosity.

Question 1. Does there exist a topology on $\mathbb{N}$ such that for all topological spaces $Y$ and all sequences $a : \mathbb{N} \rightarrow Y$ iht $a$ is continuous iff $a$ is convergent?

Now suppose we adjoin a maximum element, thereby obtaining $\mathbb{N}' = \mathbb{N} \cup \{\infty\}$.

Question 2. Does there exist a topology on $\mathbb{N}'$ sth for all Hausdorff topological spaces $Y$ and all sequences $b : \mathbb{N}' \rightarrow Y$ iht $b$ is continuous iff the restriction $a : \mathbb{N} \rightarrow Y$ is convergent and has limit equal to $b_\infty$?

Question 3. Suppose the answers are both "yes." Viewing $\mathbb{N}$ as a subset of $\mathbb{N}',$ is the topology induced on $\mathbb{N}$ the same as the topology in Question 1?

Another way to disprove (1):
Let $Y=\mathbb N$ with discrete topology. For each $n$ there is the sequence $(1,2,3,\dots,n-1,n,n,n,\dots)$ which converges to $n$. Thus we want this sequence to be continuous which implies that the preimage of every point up to $n-1$, which is just $n-1$ again, is open. So in order to make each convergent sequence continuous, $X=\mathbb N$ had to be equipped with the discrete topology. This on the other hand makes also divergent sequences continuous. — Stefan Hamcke, Apr 02 '13 at 12:12
I will abbreviate "it holds that" to "iht" and "such that" to "sth": Please don't; it may save you a little typing but it makes it a lot harder to read. — Nate Eldredge, Apr 02 '13 at 14:00

Ittay Weiss · Answer 1 · 2013-04-02T12:39:35.953

4

1) No. Suppose such a topology on $\mathbb N$ existed and consider the sequence $a_n=1/n$ as a function $a:\mathbb N \to \mathbb R$. Then $a$ is continuous. Now, consider the same function, but now change the codomain: $a:\mathbb N \to \mathbb R - {0}$. Now $a$ is not continuous. But continuity is a property of the function that is independent of the codomain, so that is impossible.

2) Yes. Metrize $\mathbb N'$ with $d(m,n)=|1/m-1/n|$ (with the convention that $1/\infty =0$). Then a sequence $a:\mathbb N\to X$, where $X$ is an arbitrary topological space, converges iff it extends to a function $a:\mathbb N'\to X$. If $X$ is Hausdorff, then limits are unique so the condition follows.

edited Apr 02 '13 at 12:39

answered Apr 02 '13 at 12:01

Ittay Weiss

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Take $Y={x}$

akkkk

Apr 02 '13 at 12:03

@akkkk, what do you mean? – Ittay Weiss Apr 02 '13 at 12:04

Oh, i'm sorry, /all/ topological spaces $Y$. I read "is there a toplogical space "Y" s.t....". My bad. – akkkk Apr 02 '13 at 12:05

Does your proof for 1) work of you replace convergent by Cauchy? The statement is then still not true... – Simon Markett Apr 02 '13 at 12:15

score 1 · Accepted Answer · answered Apr 02 '13 at 12:22

The other answers show that 1) doesn't work. Here is some "fix".

Let $O\subset \mathbb N$ be open if and only if $O=\emptyset$ or if $\mathbb N-O$ is finite. For any space $Y$ and $b\in Y$ let $Y_b$ the localisation of $Y$ at $b$, i.e. $Y=Y_b$ as sets and $O\subset Y_b$ open if and only if $O=\emptyset$ or $O\subset Y$ open with $b\in O$.

Then you have

$a:\mathbb N\to Y$ converges to $b$ if and only if $a:\mathbb N\to Y_b$ continuos.

Can convergence be seen as a form of continuity?

2 Answers2