Topology Problem: Closure X Path-Connected?

[mveritas]

Hello, I have a question about topology.

If X is a path-connected space then is it also true that closure X is path-connected?

I think it's obvious, but I can't solve it clearly...

 

[micromass]

Ah, this is a very good question. In fact, this very question was asked on my topology exam...

It is maybe obvious to you (and to me), but it is false. Consider

$$\{(x,\sin(1/x))~\vert~x\in ]0,1]\}\subseteq \mathbb{R}^2$$

This is clearly a path connected set. However, it's closure is not path connected.

I suggest that you remember this example very well. Because it is a very frequent counterexample to all sort of connectedness-properties. And it's a popular exam question as well

 

[mveritas]

Ah, this is a very good question. In fact, this very question was asked on my topology exam...

It is maybe obvious to you (and to me), but it is false. Consider

$$\{(x,\sin(1/x))~\vert~x\in ]0,1]\}\subseteq \mathbb{R}^2$$

This is clearly a path connected set. However, it's closure is not path connected.

I suggest that you remember this example very well. Because it is a very frequent counterexample to all sort of connectedness-properties. And it's a popular exam question as well

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Thank you!

 

[Landau]

It is used so often that it even has a (rather dull) name: the sine curve[/url].

 

[blue_raver22]

Dear [Name],

Thank you for reaching out with your question about topology. To answer your question, yes, it is true that if X is a path-connected space, then closure X is also path-connected. This can be proven using the definition of path-connectedness and the properties of closure in topology.

First, let's define what it means for a space to be path-connected. A space X is path-connected if for any two points x and y in X, there exists a continuous function f: [0,1] -> X such that f(0) = x and f(1) = y. In simpler terms, this means that there is a continuous path between any two points in X.

Now, let's consider the closure of X, denoted as cl(X). The closure of a set is the smallest closed set that contains all the points in the original set. In other words, it is the set of all limit points of X. We can also define it as the union of X and all its limit points.

Using these definitions, we can see that if X is path-connected, then for any two points x and y in cl(X), there exists a continuous function f: [0,1] -> X such that f(0) = x and f(1) = y. This is because cl(X) contains all the points in X, so any path between x and y in X will also be a path between x and y in cl(X). Additionally, because cl(X) is the union of X and its limit points, any path between x and y in cl(X) can also be extended to include the limit points in X.

In conclusion, the closure of a path-connected space is also path-connected. I hope this explanation helps clarify the problem for you. Feel free to reach out with any further questions or concerns.

Best,