For a function \(f:\mathbb R^n\to\mathbb R\), consider:
- \(f\) is continuous;
- the graph of \(f\) is connected in \(\mathbb R^n\times\mathbb R\).
Prove or disprove each implication \[ (1)\Longrightarrow(2), \qquad (2)\Longrightarrow(1). \]
Solution
Assume \(n\geq1\).
1. The implication \((1)\Longrightarrow(2)\) is true.
Suppose \(f:\RR^n\to\RR\) is continuous. The map \[ \Phi:\RR^n\longrightarrow\RR^n\times\RR, \qquad \Phi(x)=(x,f(x)), \] is continuous, and its image is exactly the graph \[ \Gamma_f \coloneqq \{(x,f(x)):x\in\RR^n\}. \] The space \(\RR^n\) is path connected, hence connected. Therefore its continuous image \(\Gamma_f\) is connected.
2. Define \(g:\RR\to\RR\) by\[ g(t) \coloneqq \begin{cases} \sin(1/t),&t\neq0,\\ 0,&t=0. \end{cases} \]Then \(g\) is discontinuous at \(0\).
For \[ t_m \coloneqq \frac{1}{\pi/2+2\pi m}, \] one has \(t_m\to0\) but \[ g(t_m)=1 \] for every \(m\), whereas \(g(0)=0\).
3. The graph \(\Gamma_g\) is connected.
Set \[ G_+ \coloneqq \{(t,\sin(1/t)):t>0\}, \qquad G_- \coloneqq \{(t,\sin(1/t)):t<0\}. \] Each of \(G_+\) and \(G_-\) is connected because it is the continuous image of a connected interval. Moreover, \[ \left(\frac{1}{m\pi},0\right)\longrightarrow(0,0) \] through points of \(G_+\), and \[ \left(-\frac{1}{m\pi},0\right)\longrightarrow(0,0) \] through points of \(G_-\). Hence \((0,0)\) lies in the closure of each set.
Adjoining a closure point to a connected set preserves connectedness. Therefore \[ G_+\cup\{(0,0)\} \qquad\text{and}\qquad G_-\cup\{(0,0)\} \] are connected. They meet at \((0,0)\), so their union is connected. That union is exactly \(\Gamma_g\).
4. For every \(n\geq1\), there is a discontinuous\(f:\RR^n\to\RR\) whose graph is connected.
2. The restriction of \(f\) tothe first coordinate axis is \(g\), so step 2 shows that \(f\) isdiscontinuous at the origin.
The map \[ \Gamma_g\times\RR^{n-1} \longrightarrow \Gamma_f, \qquad ((t,g(t)),y_2,\ldots,y_n) \longmapsto (t,y_2,\ldots,y_n,g(t)), \] is a homeomorphism. By step 3, \(\Gamma_g\) is connected, and \(\RR^{n-1}\) is connected. Their product is therefore connected, so \(\Gamma_f\) is connected.
5. The implication \((2)\Longrightarrow(1)\) is false.
The function constructed in step 4 has connected graph but is not continuous.
6. Q.E.D.
Step 1 proves the first implication, and step 5 disproves the second.