# Equivalent conditions on a metric space

In summary, the following are equivalent:1. A is dense in X2. The only closed set containing A is X3. The only open set disjoint from A is the empty set

## Homework Statement

Let X be a metric space and A a subset of X. Prove that the following are equivalent:
i. A is dense in X
ii. The only closed set containing A is X
iii. The only open set disjoint from A is the empty set

N/A

## The Attempt at a Solution

I can prove that i implies ii: assume that there is a closed set B containing A other than X, and show that B must equal X since the closure of A is X.

Presumably I can prove the other implications (ii -> iii and iii -> i) in a similar way but I'm not sure how to get started. Is there a certain property or fact I should be using?

So I take it that your definition of dense is "the closure is the entire space"?

OK, but what is the definition of closure?

Yes, that's the definition of dense I'm using.

The definition of closure of A, as far as I know, is the union of A with its limit points.

Good, now what is a limit point??

I want you to unwind each definition until we see something we can use. Something with open sets and with disjointness...

Ok, so a limit point of A is a point x in X such that every open ball centred at x contains another point of A, right?

I'm not sure where this is going.

OK, so a set A is dense iff all points not in A are limit points, right??

Now, take a ball disjoint from A. Are the points in this ball limit points??

micromass said:
OK, so a set A is dense iff all points not in A are limit points, right??

Right.

So if you take an open ball disjoint from A, all the points in the ball are limit points... right?

Right.

So if you take an open ball disjoint from A, all the points in the ball are limit points... right?

Yes. So can the ball be disjoint from A?? The center of the ball is a limit point, so by definition...

So it contains points of A, and therefore can't be disjoint from A. I see.

But shouldn't I be proving that ii -> iii and iii -> i?

So it contains points of A, and therefore can't be disjoint from A. I see.

But shouldn't I be proving that ii -> iii and iii -> i?

Right. But it's maybe easier to prove ii -> i first. This is extremely easy.

Now, iii -> i shouldn't be too hard if you unwind all the definitions.

Ok, I'll see how it goes.

My/your/our reasoning proves i -> iii doesn't it? The only problem is the question talks about "open sets", but we were talking about (open) balls. Or does the fact that there are no balls mean that there are no open sets? I'm a bit confused about that

Ok, I'll see how it goes.

My/your/our reasoning proves i -> iii doesn't it? The only problem is the question talks about "open sets", but we were talking about (open) balls. Or does the fact that there are no balls mean that there are no open sets? I'm a bit confused about that

Aha, yes. I forgot to mention that. You're paying attention!

Indeed, we only proved now that there are no open balls disjoint from A. But we wish to prove it for open sets.

Therefore, we will need to unwind some definitions again. What is an open set?

Eventually, I want to show that each nonempty open set contains an open ball.

Ok, well my definition of an open set is a set in which every point is an interior point.

I think I see now. If there is a nonempty open set, it contains an interior point, and thus there is an open ball within the set, by definition of interior point. So if there are no open balls there can be no open sets. Right?

Ok, well my definition of an open set is a set in which every point is an interior point.

I think I see now. If there is a nonempty open set, it contains an interior point, and thus there is an open ball within the set, by definition of interior point. So if there are no open balls there can be no open sets. Right?

Right! That is perfect!

Cool, thanks for your help! I think I've worked out the other parts now.

## 1. What is a metric space?

A metric space is a mathematical concept used to study the properties of distance and convergence. It is a set of objects with a defined distance function that satisfies certain properties, such as non-negativity, symmetry, and the triangle inequality.

## 2. What are equivalent conditions on a metric space?

Equivalent conditions on a metric space refer to different ways of characterizing or defining a metric space. These conditions may include properties such as completeness, compactness, or separability, which can be used to determine if a given set is a metric space.

## 3. How are equivalent conditions on a metric space useful?

Equivalent conditions on a metric space are useful because they provide different perspectives on a metric space, allowing for a deeper understanding of its properties and behavior. They also allow for the comparison and classification of different metric spaces.

## 4. What are some examples of equivalent conditions on a metric space?

Examples of equivalent conditions on a metric space include the Bolzano-Weierstrass theorem, which states that a subset of a metric space is compact if and only if it is sequentially compact, and the Heine-Borel theorem, which states that a subset of a metric space is compact if and only if it is closed and bounded.

## 5. How do equivalent conditions on a metric space relate to other mathematical concepts?

Equivalent conditions on a metric space have connections to other mathematical concepts, such as topology, analysis, and geometry. For example, compactness in a metric space is related to compactness in topology, and completeness in a metric space is related to completeness in analysis.

Replies
4
Views
304
Replies
2
Views
927
Replies
1
Views
1K
Replies
2
Views
1K
Replies
10
Views
2K
Replies
5
Views
4K
Replies
11
Views
2K
Replies
1
Views
1K
Replies
3
Views
2K
Replies
7
Views
1K