Let $C \subset \mathbb R^n$ a connected set. Show that if $x$ is a limit point of $C$, then $C \cup {x}$ is connected.

I first assumed that $S = C \cup {x}$ is disconnected, then, by definition, exist two subsets $U,V \subset \mathbb R^n$ such that i) $$S \subset U \cup V$$ ii) $$S \cap U \neq \emptyset$$ and $$S \cap V \neq \emptyset$$ iii) $$S \cap U \cap V = \emptyset$$

If $C \subset U$, then $x \in V$ given that $S \cap V \neq \emptyset$ and $ V \cap C = V \cap (C \cap U) \subset S \cap U \cap V = \emptyset$, but then $V$ would be a neighborhood of $x$, then $V \cap C \neq \emptyset$ given that $x$ is a limit point of C, this means that $\exists y \in V \cap C = V \cap U \cap C = \emptyset$ wich is a contradiction.

Finally, S is connected

So, ineed a review of this proof if is there something worng with it, if anyone has any suggestion or comment that would be really nice.

  • $\begingroup$ It's good .. & you got to the topological heart of it. In general, (i), If $C$ is a connected dense subspace of a space $D$ then $D$ is connected, which you prove by almost exactly the same method. (ii), So if $C$ is a connected dense subspace of $D,$ and $C\subset E\subset D$ then $E$ is connected, because $C$ is also a connected dense subspace of $E.$ In this Q apply (i) with $ D=C\cup \{x\}$. $\endgroup$ May 17 '18 at 16:14

The OP's proof is OK, but they need to explain where they are going when they write "If $C \subset U$ ...".

For the OP to understand and 'nail down' their own proof, they need to be keep in mind the distinction between a proof by contradiction and a proof by contrapositive; read on this site Proof by contradiction vs Prove the contrapositive.

The OP might find that the following crystallizes these ideas.

Premise: Let $C \subset \mathbb R^n$ and suppose $x \notin C$ with $x$ a limit point of $C$.
If $C \cup \{x\}$ is not connected then $C$ is not connected.

The OP sets everything up with (i) thru (iv), so let us take it from there with a conceptual sketch of the proof:

Assume that $x \in U$. Then $V$ must contain some points of $C$, i.e. $C \cap V \ne \emptyset$. Also, since $x$ is a limit point, $C \cap U \ne \emptyset$. So we must have

$\tag 1 C \cap V \ne \emptyset$


$\tag 2 C \cap U \ne \emptyset$

But then the open sets $U$ and $V$ used to 'disconnect' $C \cup \{x\}$ also work to show that $C$ itself is disconnected.

The same conclusion is reached if $x \in V$, so $C$ can't be connected.


You proof is fine.

There are however two enhancement that I would suggest:

  1. $C \cup x$ is not a proper writing. You should write $C \cup \{x\}$.
  2. I would not proceed supposing that $X \cup \{x \}$ is disconnected. I would suppose that $U, V$ is an open cover of $X \cup \{x\}$ and prove that either $U$ or $V$ is empty. I prefer not proceeding by contradiction when I can.

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