28

Here is an example for $\mathbb R^2 \to \mathbb R$: $$f(x,y) = \begin{cases} y & \text{when }x=0\text{ or }x=1 \\ x & \text{when }x\in(0,1)\text{ and }y=0 \\ 1-x &\text{when }x\in(0,1)\text{ and } y=x(1-x) \\ x(1-x) & \text{when }x\notin\{0,1\}\text{ and } y/x(1-x) \notin\mathbb Q \\ 0 & \text{otherwise} \end{cases} $$ This is easily ...


26

Check out this paper: F. B. Jones, Totally discontinuous linear functions whose graphs are connected, November 23, (1940). Abstract: Cauchy discovered before 1821 that a function satisfying the equation $$ f(x)+f(y)=f(x+y) $$ is either continuous or totally discontinuous. After Hamel showed the existence of a discontinuous function, many ...


10

There is a simple general strategy for many questions of this type, which is to just try to build a counterexample by transfinite induction. Let's first think about what it means for the graph $G$ of a function $f:\mathbb{R}\to\mathbb{R}$ to be disconnected. It means there are open sets $U,V\subset\mathbb{R}^2$ such that $U\cap G$ and $V\cap G$ are both ...


2

If the space is such that only constant functions are continuous (space with trivial topology), then the only path-connected sets are singletons, which are convex.


1

The comb space $C$ is not locally connected at all of its points. In fact, one can show that it is locally connected precisely at the points of $C \setminus \{0\}\times (0,1]$. The claim here is that $D' = D \cup \{(0,1/n) \mid n \in \mathbb Z \}$ is locally connected at $(0,0)$ but not locally path connected at $(0,0)$. The second part is obvious: No ...


1

Let's go with your definition of connectedness. Suppose it is connected, let there be a polygon line $\gamma$ connected two arbirarily chosen points $a,b$ in $|z|<1$ and $|z-2|<1$ respectively. Re-interpret $\gamma$ as a continuous mapping $$\gamma:[0,1]\to\mathbb{C}$$ starting at $a$ and terminating at $b$ (i.e., $\gamma(0)=a,\ \gamma(1)=b$). Let $$...


1

To prove $1 \implies 2$, you use the equation $0 = \int_\gamma \omega = f(\gamma(1)) - f(\gamma(0))$ to prove that $f$ is locally constant on $M$, because any two points in a coordinate neighborhood of $M$ are endpoints of some smooth path. Since $M$ is connected, it follows that $f$ is globally constant. Hence $x^2+y^2+z^2$ is constant on $M$, so $M$ is ...


1

Hint. Since $X$ is compact and since, as you observed, the connected component form an open cover of $X$, one can extract from them a finite subcover. On the other hand, the connected component form a partition of $X$. Thus...


1

Ferra's comment is spot-on. Let $Y = \{0\}$. Then for any space $X$ (such as $X = \{1, 2\}$), there's exactly one function from $X$ to $Y$, namely the one defined by: $$ f : Y \to X : y \mapsto 0. $$ Your claim would then show that every space is connected (including my example $X$), which is not true. Hence the claim must be false.


1

It is the union of the disjoint open sets $\{(x,y): x<0, y>0\}$ and $\{(x,y): x>0, y<0\}$. So by definition of connectedness the set is disconnected.


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