Connectedness and Continuity in Metric Spaces: Theorems and Proofs for Deeper Understanding
Explore key theorems on connectedness and continuity in metric spaces, including image of connected sets, intermediate value property, and their proofs.
Continuous Image of a Connected Metric Space
Let \( (\mathbf{X}, d_{\mathbf{X}}) \) and \( (\mathbf{Y}, d_{\mathbf{Y}}) \) be two metric spaces with \(\mathbf{X}\) be connected. And let \(f:\mathbf{X} \mapsto \mathbf{Y}\) be function. If \(f\) is continuous then \( f\left(\mathbf{X} \right)\) is connected metric space induced from \(Y\).
Given that \( (\mathbf{X}, d_{\mathbf{X}}) \) and \( (\mathbf{Y}, d_{\mathbf{Y}}) \) are two metric spaces with \(\mathbf{X}\) is connected.
Let \(f:\mathbf{X} \mapsto \mathbf{Y}\) is continuous.
To prove \( f\left(\mathbf{X} \right)\) is connected.
If possible let \( f\left(\mathbf{X} \right)\) is disconnected. Then there exists a continuous mapping \(g:f\left(\mathbf{X} \right) \mapsto \mathbf{X_0} \) which is onto and \( (\mathbf{X_0},d_{0}) \) is a discrete two-point metric space \( (\mathbf{X_0},d_{0}) \) where \( \mathbf{X_0} = \{0,1\} \).
Since \(f\left(\mathbf{X} \right)\subseteq \mathbf{Y} \) then \(g\circ f: \mathbf{X} \mapsto \mathbf{X_0}\) exists.
Again \(f\) and \(g\) are both continuous then \(g\circ f\) is continuous. This implies \(\mathbf{X} \) is disconnected. A contradiction.
Therefore our assumption is wrong.
Hence \( f\left(\mathbf{X} \right)\) is connected.
Intermediate Value Theorem for Continuous Function on [0,1]
Let \(f:[0,1]\mapsto \mathbb{R}\) is continuous then for every \(y\) such that \(f(a)\leq y\leq f(b)\) or \(f(b)\leq y\leq f(a)\) there exists \(x\in[a,b]\) for which \(f(x)=y\).
Given that \(f:[0,1]\mapsto \mathbb{R}\) is continuous.
Let \(f(a)\leq y\leq f(b)\).
Case A: \(f(a)= y\)
Then \(x=a\) for which \(f(x)=y\).
Case B: \(f(b)= y\)
Then \(x=b\) for which \(f(x)=y\).
Case C: \(f(a)\lt y \lt f(b)\)
Since \([0,1]\) is an interval in \(\mathbb{R}\) then \([0,1]\) is connected.
Since \(f\) is continuous and \([0,1]\) is connected then \(f\left([0,1] \right)\) is connected.
Since \(f\left([0,1] \right)\subseteq \mathbb{R} \) and \(f\left([0,1] \right)\) is connected then \(f\left([0,1] \right)\) is interval.
Since \(f\left([0,1] \right)\) is interval, \(f(a),f(b) \in f\left([0,1] \right) \) and \(f(a)\lt y \lt f(b) \) then \(y\in f\left([0,1] \right)\).
Then there exists a preimage \(x\in [0,1]\) such that \(f(x)=y\). Hence the theorem is proved.
Characterization of Connectedness via Intermediate Value Property
Let \( (\mathbf{X}, d_{\mathbf{X}}) \) be a metric space. If every continuous function \(f:(\mathbf{X}, d_{\mathbf{X}}) \mapsto (\mathbb{R},d)\) has the intermediate value property (i.e., if \(y_{1},y_{2}\in f\left(\mathbf{X} \right)\) and \(y\in \mathbb{R}\) such that \(y_{1}\lt y\lt y_{2}\) then there exists an \(x\in \mathbf{X} \) such that \(f(x)=y\) ) then \(\mathbf{X}\) is connected.
Given that \( (\mathbf{X}, d_{\mathbf{X}}) \) is a metric space.
Let every continuous function \(f:(\mathbf{X}, d_{\mathbf{X}}) \mapsto (\mathbb{R},d)\) has the intermediate value property.
To prove \(\mathbf{X}\) is connected.
If possible let \(\mathbf{X}\) is disconnected. Then there exists a continuous mapping \(g:(\mathbf{X}, d_{\mathbf{X}}) \mapsto (\mathbf{X_{0}}, d_{0})\) which is onto.
Let us construct a mapping \(h:(\mathbf{X_{0}}, d_{0}) \mapsto (\mathbb{R},d)\) such that \(h(0)=0 \) and \(h(1)=1 \). Clearly \(h\) is continuous.
Clearly, \(h\circ g:(\mathbf{X}, d_{\mathbf{X}}) \mapsto (\mathbb{R},d)\) exists. Since \(g\) and \(h\) are continuous then \(h\circ g\) is also continuous. Also, \(0\in (h\circ g)(\mathbf{X})\) and \(1\in (h\circ g)(\mathbf{X})\).
Since \(0\lt \frac{1}{2} \lt 1\) then there exists no \(x\in \mathbf{X} \) such that \(f(x)=\frac{1}{2}\).
So, \(h\circ g\) does not have the intermediate value property. A contradiction.
Therefore our assumption is wrong.
Hence \(\mathbf{X}\) is connected.