# The closure of a connected set is connected

• librastar
In summary, the conversation discusses proving that the closure of a connected subset in a topological space is also connected. The attempt at a solution involves using a combination of definitions and theorems involving discrete valued maps and continuous and onto functions. There is a question about whether it is allowed to simply define a function without explicitly showing its continuity and surjectivity. It is also noted that such a function may not exist in general. The conversation ends with a request for a hint or direction on how to approach the problem.
librastar

## Homework Statement

Let X be a topological space. Let A be a connected subset of X, show that the closure of A is connected.
Note: Unlike regular method, my professor wants me to prove this using an alternative route.

## Homework Equations

a) A discrete valued map, d: X -> D, is a map from a topological space X to a discrete space D.
b) A topological space X is connected if and only if every discrete valued map on X is constant.
c) Suppose X is connected and f:X -> Y is continuous and onto. Then Y is connected.

## The Attempt at a Solution

My intuition tells me that if I can use the combination of a), b) and c) I should be able to arrive at the solution. Here is my rough idea (note that cl(A) means closure):

Let A be a connected subset of X.
Let f: A -> cl(A) be a continuous and onto function.
Take a arbitrary discrete valued map d: cl(A) -> D.
Consider (composition) d$$\circ$$f: A -> D which is a discrete valued map on X.
By b) since A is connected, d$$\circ$$f is constant.
This shows that d is constant.
Again by b), since d is a constant, this implies that cl(A) is connected.

To me, the most troubling problem is when defining such f.
Am I allowed to simply define an f that is continuous and onto, even if I do not explicit show it is continuous and onto?

Thanks.

f isn't going to exist in general. For example, if you put the indiscrete topology on N, and look at the closure of {1}, the closure has a larger cardinality even

Ok, so I cannot just define the existence of such function.

So can you give a hint or point a direction on how to work on my problem?

## 1. What is a connected set?

A connected set is a set in which every pair of points can be connected by a continuous curve within the set. In other words, there are no breaks or gaps in the set.

## 2. What does it mean for a set to be closed?

A closed set is one in which all its limit points are contained within the set. In other words, it contains all its boundary points.

## 3. What is the closure of a set?

The closure of a set is the smallest closed set that contains all the points in the original set. It includes all the boundary points of the set.

## 4. Why is the closure of a connected set connected?

This is because the closure of a set includes all the boundary points of the original set. Since a connected set has no gaps or breaks, the closure must also have no gaps or breaks, making it connected as well.

## 5. Can a disconnected set have a connected closure?

No, a disconnected set cannot have a connected closure. This is because the closure of a disconnected set would have gaps or breaks, which would make it disconnected as well.

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