1. ## Connected Set Problem

Let A be a connected subset of R^n. Let p be a limit point of A. Show that A U {p} is connected.

2. Helping with proofs involving connected sets is very difficult without knowing the exact definition and sequence of theorems in your textbook. There are many different but equivalent approaches to connected sets. This particular question is either a simple consequence of the theorem- “The closure of a connected set is connected”. Or it may be a lemma to the proof of that theorem. Basically you would show that if Au{p} is separated some open set contains p and no other point of A. That is a contradiction.

3. Well, I'm trying to solve it this way:

A is a connected set, implies A has Intermediate value property, so for each continuous function f:A->A, f(A) is an interval.

Now p is a limit point of A, so there exist a sequence {Pn} in A\{p} such that {Pn} -> p.

And I'm thinking if I can suppose to the contrary that A is not connected, then the function from {Pn} to p cannot be an interval, in which is a contradiction.

But something doesn't look right. How do I know that Pn to p is continuous?

Or should I use the seperate operators? Say there are U and V in R^n that seperates AU{p}, then I can try to get a contradiction. Oh yeah, that sounds a lot more convincing.

A is a connected set, implies A has Intermediate value property, so for each continuous function f:A->A, f(A) is an interval.
In a lot years of teaching topology, I have never seen that approach to connected sets. But in any case, because f is continuous then f(p_n)-> f(p)

Say there are U and V in R^n that seperates AU{p}, then I can try to get a contradiction.
That is the approach that I posted above.

5. But how do I know that the limit point p is inside A? All I know is within the negbourhood of p (Nr(p)), there exist at least a point from A. Should I use that fact in addition to the separators to prove this?