Prove that a sequence converges in this topological space iff

[DotKite]

Homework Statement

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

Homework Equations

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

 

[Dick]

Homework Statement

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

Homework Equations

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

Yes, that's what it means.

 

[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?

 

[Dick]

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?

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.

 

[DotKite]

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

 

[Dick]

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

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

 

[DotKite]

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

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

 

[Dick]

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

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

 

[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

 

[Dick]

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

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)?

 

[DotKite]

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

 

[Dick]

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

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.

 

[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 $\left \{ X_n \right \} \rightarrow \delta$,

 

[Dick]

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 $\left \{ X_n \right \} \rightarrow \delta$,

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.