Definition. A topological space \(X\) is disconnected if \(∃U, V\) open, disjoint, non-empty such that \(X=U∪V\). If \(X\) is not disconnected, it is called connected.
Proposition. Let \(X\) be a topological space. The following are equivalent:
Proof. Seen in Metric Spaces.◻
Definition. A non-empty subset \(A\) of a topological space is connected if \(A\) with the subspace topology is connected.
Remark. ∅ is connected by convention.
Remark. Equivalent definition: \(A\) is connected if \(∀U, V\) open in \(X\) such that \(A⊂U∪V\) and \(A∩U∩V=∅\), either \(U∩A=∅\) or \(V∩A=∅\).
Example. ℝ with cofinite topology.
\(A=\{1, 2\}\) is not connected. \(U=ℝ∖\{2\}, V=ℝ∖\{1\}, U∩A=\{1\}\) and \(V∩A=\{2\}\) are open in \(A\).
On the other hand, for every \(U, V\) open in (ℝ, cofinite) such that \(1∈U, 2∈V, U∩V≠∅\).
Proposition. If \(f:X→Y\) is continuous, \(A⊂X\) is connected, then \(f (A)\) is connected.
Proof. Seen in Metric Spaces◻
Example. \([0, 1], S^1\) are both connected but not homeomorphic. Say \(f:[0, 1]→S^1\) is a homeomorphism, then \(f|_{[0, 1]-\left\{\frac12\right\}}:\left[0, \frac12\right)∪\left(\frac12, 1\right]→S^1 ∖\left\{f \left(\frac12\right)\right\}\) is homeomorphism, contradiction since \(S^1∖\left\{f \left(\frac12\right)\right\}\) is connected but \(\left[0, \frac12\right)∪\left(\frac12, 1\right]\) is not.
Proposition. Suppose \(\{A_i:i∈I\}\) is a family of connected subsets of a topological space \(X\) such that \(A_i∩A_j≠∅ ∀i, j∈I\). Then \(⋃_{i∈I}A_i\) is connected.
Proof. Let \(f:⋃_{i∈I}A_i→\{0, 1\}\) be continuous. Since \(A_i\) is connected, \(f (A_i)=c_i∈\{0, 1\}\), since \(A_i∩A_j≠∅\), let \(x∈A_i∩A_j\), then \(c_i=f (x)=c_j\), so \(c_i=c_j ∀i, j∈I\), so \(f\) is constant, so \(⋃_{i∈I}A_i\) is connected.◻
Corollary. If \(\{C_i:i∈I\}\) and \(B\) are connected subsets of a topological space \(X\) and \(C_i∩B≠∅ ∀i\), then \(\left(⋃_{i∈I}C_i\right) ∪B\) is connected.
Proof. Take \(A_i=C_i∪B, A_i∩A_j \supset B\), so \(A_i∩A_j≠∅\).◻
Example. \(A⊂ℝ^2\)
\(A=\{(x, y):x∈ℝ, y∈\mathbb{Q}\}∪\{(0, x):x∈ℝ\}\)
By the corollary, \(A\) is connected.
Theorem. If and only if \(X, Y\) are connected, the topological product \(X×Y\) is connected.
Proof. (\(⇒\)) \(p_X:X×Y→X\) is continuous, so \(p_X (X×Y)=X\) is connected and similar for \(Y\).
(\(⇐\)) Note: For any \(y_0∈Y\) fixed, the map \(i:X→X×Y\) given by \(i (x)=(x, y_0)\) is continuous.
Proof: If \(U×V\) is open with \(y_0∉V\), \(i^{-1}(U×V)=∅\)
If \(U×V\) is open with \(y_0∈V\), \(i^{-1}(U×V)=U\) open
So \(X×\{y\}\) is connected \(∀y∈Y\).
\(B=\{x_0\}×Y\) is connected.
\(B∩(X×\{y\})=\{(x_0, y)\}≠∅\)
\(X×Y=B∪⋃_{y∈Y}(X×\{y\})\) is connected by the corollary.◻
Remark. \(X_1×⋯×X_n\) is connected iff \(X_i\) connected \(∀i\).
Theorem. Suppose \(A\) is a connected subset of a topological space \(X\) and \(A⊂B⊂\overline{A}\). Then \(B\) is connected.
Proof. Let \(U, V\) be open in \(X\) such that \(B⊂U∪V\) and \(B∩U∩V=∅\). It is enough to show that \(B∩U\) or \(B∩V\) is empty.
Since \(B∩U∩V=∅\) we have \(A∩U∩V=∅\), but \(A\) is connected, so either \(A∩U=∅\) or \(A∩V=∅\), say \(A∩V=∅\). Then \(A⊂X∖V\) closed\(⇒ \overline{A}⊂X∖V\).
\(B⊂\overline{A}⇒B⊂X∖V⇒B∩V=∅\).◻
Definition. A path \(P\) connecting 2 points \(x,y\) in \(A\) is a continuous map \(p:[0,1]→X\) such that \(p(0)=x,p(1)=y\).
Remark. We know \([0,1]\) is connected, so \(p([0,1])\) is connected.
Definition. A topological space \(X\) is path connected if \(∀a, b∈X\) there is a path connecting \(a, b\).
\(A⊂X\) is path connected if \(∀a, b∈A, ∃p:[0, 1]→A\) continuous such that \(p (0)=a, p (1)=b\).
Proposition. If \(X\) is path connected, then \(X\) is connected.
Proof. Fix \(x_0∈X\). For any \(y∈X\), take \(p_y:[0, 1]→X\) such that \(p (0)=x_0, p (1)=y\). \(P_y=p_y ([0, 1])\).
\(X=⋃_{y∈X}P_y, P_y∩P_z⊃\{x_0\}\), so \(P_y∩P_z≠∅\), so by proposition \(X\) is connected.◻
Example. The image of $⟨(1+1/t)\cos t,(1+1/t)\sin t⟩$ along with the unit circle is connected in \(ℝ^2\) but not path connected. MSE
Proposition. Let \(f:X→Y\) be continuous map of topological spaces, \(A⊂X\) path connected. Then \(f (A)\) is path connected.
Proof. If the path \(p:[0, 1]→X\) joins \(a, b\), the path \(f∘p\) joins \(f (a), f (b)\).◻
Definition. A family \(\{U_i:i∈I\}=𝒰\) of subsets is called a cover of \(X\) if \(X=⋃_{i∈I}U_i\). If each \(U_i\) is open in \(X\), \(𝒰\) is called an open cover of \(X\). A subcover of a cover \(\{U_i:i∈I\}\) is a subfamily \(\{U_i:i∈J\}\) for some \(J⊂I\) such that \(X=⋃_{j∈J}U_j\).
We say that this subcover is finite(countable) if \(J\) is finite(countable).
Definition. A topological space \(X\) is compact if every open cover of \(X\) has a finite subcover.
A subset \(A\) of \(X\) is compact if \(A\) is compact if \(A\) is compact with the subspace topology.
\(\{U_i:i∈I\}\) is a cover of \(A⊂X\) if \(A⊂⋃_{i∈I}U_i\) (Similarly for open cover, subcover)
Remark. \(A\) is compact if every open cover of \(A\) has a finite subcover.