$\forall S \subseteq \mathbb R^m, (S:\text{Connected}\land S:\text{Open in } \mathbb R^m \to S: \text{Path Connected})$

I searched on this website, and I could only find two proofs.

The first proof uses the concept of locally path connectedness, but I do not know this.

About the other proof, I cannot understand the second paragraph.

Let $x\in A$ (hence not empty); suppose $\exists \varepsilon >0 : B(x,\varepsilon)\subseteq U$. Since open balls are convex, it is path connected. Thus for any point $y$ in $B(x)$, there is a path from $x$ to $y$. Since $A$ is a set of points in $U$ that can join $a$, then there exist a path between $x$ and $a$. Since there is a path from $y$ to $x$ and $x$ to $a$; then, there is a path from $y$ to $a$ implying $y$ is in $A$. Since $y\in B(x,\varepsilon)$, we conclude $B(x) \subseteq A$, hence $A$ is open

Could you provide the easy proof?

  • $\begingroup$ See this. $\endgroup$ Jul 23, 2017 at 22:56
  • $\begingroup$ What exactly don't you understand? $\endgroup$ Jul 23, 2017 at 23:02

1 Answer 1


Do you know the basic fact that is mentioned that if there is a path from $x$ to $y$ and another path from $y$ to $z$ we can construct a combined path from $x$ to $z$? That is the main idea, plus that we have a (straight line) path from the centre of a ball to any of its points, by convexity of balls.

  • $\begingroup$ Because of your comment, I finally could understand the second paragraph!! I totally understand now. Thank you so much! $\endgroup$
    – acubens555
    Jul 24, 2017 at 2:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .