# Proof check: If $X$has more than 2 points and $f$ is a one to one in $\mathbb{R}$, then $\exists x \in X$ s.t. $X \setminus \{x\}$ is disconnected.

Proof check: If $X$ is a topological space with more than 2 points and $f$ is a one to one function to $\mathbb{R}$, then $\exists x \in X$ s.t. $X \setminus \{x\}$ is disconnected.

Proof: $X$ has at least 3 points, $x_1, x_2, x_3$ that map to $y_1 < y_2 < y_3$. Assume that $X \setminus \{x_2\}$ is connected and therefore path connected. Therefore there exists $\gamma(t): [0,1] \to \mathbb{R}$, $\gamma(0) = x_1,\gamma(1) = x_3$. Define $g:= f \circ \gamma$ a continous function in $\mathbb{R}$, therefore there exists $x$ s.t. $\gamma (t) = x, f(x) = y_2$, which is of course a contradiction to $f$ being one to one.

I'm pretty sure this proof is correct (is it?) but it seems like there has to be a simpler way of proving this, perhaps only by using connectedness and not path connectedness. Is there?

• You should state that $f$ is continuous. – J.-E. Pin Sep 17 '15 at 8:36
• connectedness does not imply path-connectedness. – drhab Sep 17 '15 at 8:39

Your proof is not correct:

Your first mistake (and most important one) is:

Assume that $X\setminus \{x_2\}$ is connected and therefore path connected.

False. There exist connected spaces that are not path connected. The implication works the other way around.

Second mistake:

Therefore there exists $\gamma(t):[0,1]\to\mathbb R$, $\gamma(0)=x_0,\gamma(1)=x_3$.

Since $\gamma$ maps to $\mathbb R$ and $x_0$ is an element of $X$, this makes no sense.

Hints:

• If $f:X\rightarrow Y$ is continuous and $Y$ is not connected then $X$ is not connected.

• If moreover $f$ is injective then the function $g:X-\{x_0\}\rightarrow Y-\{f(x_0)\}$ prescribed by $x\mapsto f(x)$ is well-defined and continuous.

• If $y_0\in\mathbb R$ then $\mathbb R-\{y_0\}$ is not connected.