Connectedness & Compactness¶
Connectedness¶
Connectedness defined by separations
Suppose \(X\) is a topological space. A separation of \(X\) is a pair \(U\), \(V\) of disjoint non-empty open subsets of \(X\), where \(X=U\cup V\).
The space \(X\) is said to be connected, if there does not exist a separation of \(X\).
Equivalent definition of connectedness
The space \(X\) is connected, iff the only subsets of \(X\) that are both open and closed in \(X\) are \(\varnothing\) and \(X\) itself.
We show that for a non-empty subset \(X\), it could not be both open and closed.
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Necessary. If a non-empty subset \(A\) of \(X\) is both open and closed, then \(B=X-A\) is open and \(X=A\cup B\), so \(A\), \(B\) are a separation of \(X\).
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Sufficient. If \(U\) and \(V\) is a separation of \(X\), then \(U\) itself is both open and closed.
\(\square\)
Example. Show that \(\mathbb{E}^1\) is connected.
Properties of connectedness¶
Separation and connected subspace
Suppose \(U\subset X\) is both open and closed, \(A\subset X\) is connected, then either \(U\cap A = \varnothing\) or \(U\cap A=A\), i.e. \(A\subset U\).
The above lemma could be stated as follows. Suppose \(X=U\cup V\), where \(U\) and \(V\) is a separation of \(X\), then for a connected subspace \(A\subset U\), it lie entirely either within \(U\) or \(V\).
By definition.
Properties of connectedness
(i) The continuous image of connected space is connected.
(ii) Suppose \(A\subset X\) is connected, and \(A\subset Y\subset \overline{A}\), then \(Y\) is connected.
(iii) Suppose \(\mathcal{U}\) is a connected covering of \(X\), \(A\subset X\) is connected, and \(A\cap U\neq \varnothing\) for all \(U\in\mathcal{U}\), then \(X\) is connected.
Example. Using (iii) to show that
(i) \(\mathbb{E}^2\) is connected.
(ii) \(S^n\) is conncected.
Path connectedness¶
path, path connectedness
Given points \(x,y\in X\), a path in \(X\) from \(x\) to \(y\) is a continuous map \(f:[a,b]\rightarrow X\) of some closed interval in the real line into \(X\), such that \(f(a)=x\), \(f(b)=y\). Specially we choose \(a=0,b=1\).
\(X\) is said to be path connected, if every pair of points of \(X\) can be joined by a path in \(X\).
Properties of path connectedness
(i) A path connected space is connnected.
(ii) Continuous image of a path connected image is path connected.
The following example shows that connected space is necessarily a path connected space.
Example. Topologist’s sine curve. Suppose
show that
(i) \(A\) is connected, so \(\overline{A}=A\cup B\) is connected.
(ii) \(\overline{A}\) is not path connected.
Compactness¶
Coverings
A collection \(\mathcal{A}\) of subsets of a space \(X\) is said to cover \(X\), or to be a covering of \(X\), if the union of elements of \(\mathcal{A}\) equals \(X\), i.e.
If all elements of \(\mathcal{A}\) is open, then \(\mathcal{A}\) is called an open covering of \(X\).
Compact set
Set \(X\) is said to be compact, if any open covering of \(X\) has finite-number subcollection that also covers \(X\).
Now we give some properties about compactness.
compactness of subsets
A collection \(\mathcal{U}\) of \(X\), is said to be a covering of \(A\) in \(X\), if \(A \subset \bigcup_{U\in \mathcal{U}}U\).
Suppose \(A\subset X\). \(A\) is compact, iff any open covering of \(A\) in \(X\) has finite subcollection that also covers \(A\).
Compactness is closely connected with the concept of closedness. When does a closed set be a compact set?
Compactness for closed set
Every closed subspace of a compact space is compact.
Like we have in continuous analysis, we have the following theorem.
Compactness of Image of compact set under a continuous map
The image of a compact space under a continuous map is compact.
Compact subspace of Hausdorff space¶
This is the result of compactness and \(T_2\) axiom.
closedness of compact subspace of Hausdorff space
Every compact subspace of a Hausdorff space is closed.
That is, for a compact subspace \(A\) of a Hausdorff space \(X\), choose \(x\not\in A\), then there exsit disjoint open set \(U\) and \(V\), such that \(x\in U\) and \(Y\subset V\).
We give a proof by showing that for all \(x\in A^c\), \(x\) has a neighborhood \(U\) such that \(U\cap A=\varnothing\), i.e. \(x\in U\subset A^c\), so \(A^c\) is open, so \(A\) is closed.
Notice for all \(y\in A\), since \(X\) is Hausdorff, there exists open set \(U_y\) and \(V_y\) s.t. \(x\in U_y\), \(y\in V_y\) and \(U_y\cap V_y=\varnothing\). Also notice that \(\bigcup_{y\in A}V_y\) is an open covering of \(A\). Since \(A\) is compact, there exsits finite number \(N\) such that \(Y\subset \bigcup_{i=1}^N V_{y_i}\). Let
which are still open sets, and \(U_x\cap A=\varnothing\). \(U_x\) is the open set that we want.
\(\square\)
The following theorem is a useful tool to show whether a map is a homeomorphism. The proof is done by transfering closedness and compactness.
A sufficient condition for a homeomorphism
Assume \(f:X\rightarrow Y\) is a continuous bijection. If \(X\) is compact and \(Y\) is a Hausdorff space, then \(f\) is a homeomorhpism.
We only need to prove that \(f^{-1}\) is continuous. By relationship of open, closedness and continuous, we suffice to show that \(f\) is closed.
For a closed subspace \(A\) of \(X\), since \(X\) is compact, by Compactness for closed set, we have \(A\) is compact. So is \(f(A)\) in \(Y\) by Compactness of Image of compact set under a continuous map. Since \(Y\) is Hausdorff, then by closedness of compact subspace of Hausdorff space, we have \(f(A)\) is closed.
\(\square\)
compact Hausdorff space satisfies T3, T4 axioms
Compact Hausdorff space satisfies \(T_3\), \(T_4\) axioms.
Suppose \(X\) is a Hausdorff space. We suffice to show that \(X\) satisfies \(T_4\) axiom. Choose two disjoint closed subspace \(A\), \(B\) of \(X\), by Compactness for closed set, \(A\), \(B\) are two disjoint compact spaces.
\(\square\)
Compactness on product space¶
Tube Lemma
Consider product \(X \times Y\), where \(X\) is compact. Choose \(y\in Y\). Assume \(N\) is an open set of \(X\times Y\) that contains the slice \(X\times \{y\}\), then \(N\) contains some tube \(X\times W\), where \(W\) is a neighborhood of \(\{y\}\) in \(Y\).
productivity of compactness
The product of finitely many compact spaces is compact.
We only prove for the case of two compact spaces.