# Prove that a sequence converges in this topological space iff

1. Dec 8, 2013

### DotKite

1. The problem statement, all variables and given/known data

Consider (R,C). Prove that a sequence converges in this topological space iff it is bounded below
define $C =$ $\left \{ (a,\infty)|a\in R \right \} \bigcup \left \{ \oslash , R \right \}$

2. Relevant equations

3. The attempt at a solution

So I am not very clear on how to wrap my head around what (R,C) actually means. I know that R is the underlying set and that C is the topology. So does that mean that, in this case, that all open sets in the underlying set, R, are of this form (a,∞)? And does that mean any open ball around a point in R is of that same form? Please help

2. Dec 8, 2013

### Dick

Yes, that's what it means.

3. Dec 8, 2013

### DotKite

OK so if we let $\left \{ X_n \right \} \rightarrow x$, it follows that for each neighborhood $V$ of x, $\exists N_0 \in \mathbb{N}$ s.t $X_n \in V$ for $n > N_0$.

Since $V$ is a neighborhood of x it contains an open set, call it $O$ that contains x.

Here I get stuck. Does this mean that $\left \{ X_n \right \}$ is in $O$ for $n > N_0$ and $O$ is bounded below by x?

Last edited: Dec 8, 2013
4. Dec 8, 2013

### Dick

You are kind of taking it the wrong way, you don't even know $X_n$ has a limit yet. Suppose $X_n$ is bounded below by a number $B$. What kinds of points do you think might be candidates for the limit? You know what the neighborhoods look like. Try to think in terms of pictures rather than writing a bunch of symbols down first.

5. Dec 8, 2013

### DotKite

arent you suppose to assume a sequence converges for the => direction?

6. Dec 8, 2013

### Dick

Yes, you are. I was thinking of the opposite direction. So ok, what are all the neighborhoods that can contain x?

7. Dec 8, 2013

### DotKite

neighborhoods that can contain x,
$(a,\infty)$ where a<x?

8. Dec 8, 2013

### Dick

Exactly. So $(x-1,\infty)$ contains all but a finite number of the $X_n$. Does that show $X_n$ is bounded?

9. Dec 8, 2013

### DotKite

well we are suppose to show that $X_n$ is bounded below. I suppose it would be bounded below by x-1 if it is increasing, and bounded above and below if it is decreasing? The neighborhood going towards infinity is confusing me. I guess if it is decreasing it can't start at infinity it has to start at a point, and it would stay in $(x-1,\infty)$ and be bounded above by x-1

10. Dec 8, 2013

### Dick

You are probably over thinking this. There is some N such that $(x-1,\infty)$ contains all $X_n$ with n>N. So those points are all bounded below by x-1. The points you haven't bounded yet is the finite collection $X_1, X_2, ... X_n$. Are those bounded below (not necessarily by x-1)?

11. Dec 8, 2013

### DotKite

oh i see. So take that finite collection and take the minimum, then that finite collection will be bounded below by that min?

12. Dec 8, 2013

### Dick

Sure, so the whole sequence is bounded below by the minimum of that min and x-1. Now try the other direction. It's actually even easier.

13. Dec 8, 2013

### DotKite

ok, so we assume that there is a sequence bounded bellow by some number. Call it $\delta$. Thus $X_n$ > $\delta$ $\forall n$. Let ε > 0. $N_\delta$ = $\left ( \delta - ε, \infty \right )$ is the set of all neighborhoods of $\delta$. Notice that $X_n \in N_\delta$ $\forall n$ since $X_n$ > $\delta - ε$ $\forall n$. Thus $\left \{ X_n \right \} \rightarrow \delta$,

14. Dec 8, 2013

### Dick

You've got the concept fine. Your notation is a little lacking. Having a lower bound δ means $X_n \ge \delta$. Don't you mean to define $N_\epsilon=\left ( \delta - ε, \infty \right )$? Try and rewrite that a bit. And what's maybe interesting about this topology is that $\left \{ X_n \right \} \rightarrow \delta-1$ is also true.

Last edited: Dec 8, 2013