One session on point-set topology, with problems on separation axioms and compactness. The workshop packets are listed under topology workshop packets.
Product, subspace, and quotient topologies (Tag E-YUGKU)
- State the definition of the product topology, the subspace topology, and the quotient topology.
Solution
<1>1. The product topology on \(X\times Y\) is the topology with basis \(\{U\times V:U\subseteq X\text{ open},\ V\subseteq Y\text{ open}\}\).
This is the defining basis for the product topology.
<1>2. If \(A\subseteq X\), the subspace topology on \(A\) is \(\{A\cap U:U\subseteq X\text{ open}\}\).
This is the definition of the topology induced by the inclusion \(A\hookrightarrow X\).
<1>3. If \(q:X\twoheadrightarrow Y\) is a surjection, the quotient topology on \(Y\) is defined by \[U\subseteq Y\text{ open}\iff q^{-1}(U)\subseteq X\text{ open}.\]
This is the definition of the quotient topology induced by \(q\).
Definition of a basis for a topology (Tag E-OIS5D)
- What does it mean for a family of sets to form a basis for a topology?
Solution
<1>1. A family \(\mathcal B\) of subsets of \(X\) is a basis if every point of \(X\) lies in some \(B\in\mathcal B\) and whenever \(x\in B_1\cap B_2\) with \(B_1,B_2\in\mathcal B\), there is \(B_3\in\mathcal B\) with \(x\in B_3\subseteq B_1\cap B_2\).
These two conditions are precisely the basis axioms.
<1>2. The topology generated by \(\mathcal B\) consists of all unions of elements of \(\mathcal B\).
The basis axioms ensure these unions contain \(\varnothing,X\) and are closed under arbitrary unions and finite intersections.
Interior, isolated, and limit points (Tag E-EE4EE)
- What is an interior point? An isolated point? A limit point?
Solution
<1>1. Let \(X\) be a topological space and \(A \subseteq X\).
setup.
<1>2. A point \(x \in A\) is an interior point of \(A\) if there is an open set \(U\) with \(x \in U \subseteq A\).
definition of interior point.
<1>3. A point \(x \in A\) is an isolated point of \(A\) if there is an open set \(U\) with \(U \cap A = \{x\}\).
definition of isolated point.
<1>4. A point \(x \in X\) is a limit point (accumulation point) of \(A\) if every open set \(U\) containing \(x\) meets \(A\) in a point other than \(x\), i.e. \((U \setminus \{x\}) \cap A \neq \varnothing\).
definition of limit point.
<1>5. Q.E.D.
<1>2, <1>3, <1>4.
Closure of a subspace (Tag E-KJQUD)
- What is the closure of a subspace \(E\subseteq X\)?
Solution
<1>1. The closure \(\overline E\) is the intersection of all closed subsets of \(X\) containing \(E\).
Arbitrary intersections of closed sets are closed, so this is the unique smallest closed subset containing \(E\).
<1>2. Equivalently, \[\boxed{\overline E=\{x\in X:U\cap E\ne\varnothing\text{ for every open neighborhood }U\ni x\}.}\]
A point fails to lie in the intersection from <1>1 exactly when it has an open neighborhood contained in the complement of some closed set containing \(E\), hence disjoint from \(E\).
Compact topological space (Tag E-FLVZU)
- What does it mean for a topological space to be compact?
Solution
<1>1. A topological space \(X\) is compact if every open cover of \(X\) has a finite subcover.
Explicitly, whenever \(X=\bigcup_{i\in I}U_i\) with each \(U_i\) open, there exist \(i_1,\dots,i_r\in I\) such that \(X=U_{i_1}\cup\cdots\cup U_{i_r}\). This is the definition of compactness.
Dense subspace (Tag E-6W4WA)
- What does it mean for \(E\subseteq X\) to be a dense subspace?
Solution
A subset \(E\subseteq X\) is dense in \(X\) if \[ \overline E=X. \] Equivalently, every nonempty open subset of \(X\) meets \(E\).
A point \(x\) lies in \(\overline E\) exactly when every open neighborhood of \(x\) intersects \(E\). Hence \(\overline E=X\) exactly when every nonempty open set intersects \(E\).
Six topologies on $\mathbb{R}^{d}$ (Tag E-XSXAZ)
Construct 6 distinct topologies on the set \(\mathbb{R}^d\) (\(d \ge 1\)), and order them by coarseness/fineness where applicable.
Solution
Let \(S=\mathbb R^d\).
<1>1. The following are six distinct topologies on \(S\): <2>1. The indiscrete topology \[ \mathcal T_{\mathrm{ind}}=\{\varnothing,S\}. \] <2>2. The cofinite topology \[ \mathcal T_{\mathrm{cof}}=\{\varnothing\}\cup\{U\subseteq S:S\setminus U\text{ is finite}\}. \] <2>3. The cocountable topology \[ \mathcal T_{\mathrm{coc}}=\{\varnothing\}\cup\{U\subseteq S:S\setminus U\text{ is countable}\}. \] <2>4. The Euclidean topology \(\mathcal T_{\mathrm{eucl}}\). <2>5. The \(d\)-fold lower-limit (Sorgenfrey product) topology \(\mathcal T_{\mathrm{Sorg}}\), with basis \[ \prod_{j=1}^d[a_j,b_j),\qquad a_j<b_j. \] <2>6. The discrete topology \[ \mathcal T_{\mathrm{disc}}=\mathcal P(S). \]
<1>2. The following strict inclusions hold: \[ \mathcal T_{\mathrm{ind}} \subsetneq\mathcal T_{\mathrm{cof}} \subsetneq\mathcal T_{\mathrm{coc}} \subsetneq\mathcal T_{\mathrm{disc}}, \] and \[ \mathcal T_{\mathrm{ind}} \subsetneq\mathcal T_{\mathrm{cof}} \subsetneq\mathcal T_{\mathrm{eucl}} \subsetneq\mathcal T_{\mathrm{Sorg}} \subsetneq\mathcal T_{\mathrm{disc}}. \] <2>1. Every cofinite open set is Euclidean-open because finite subsets of \(\mathbb R^d\) are Euclidean-closed. <2>2. Every Euclidean-open set is Sorgenfrey-open: each Euclidean-open rectangle is a union of half-open basic rectangles. <2>3. Strictness of \(\mathcal T_{\mathrm{eucl}}\subsetneq\mathcal T_{\mathrm{Sorg}}\) is witnessed by \([0,1)\times\mathbb R^{d-1}\).
<1>3. The Euclidean and cocountable topologies are incomparable. <2>1. A bounded Euclidean open ball is not cocountable, since its complement is uncountable. <2>2. The set \[ S\setminus\mathbb Q^d \] is cocountable because \(\mathbb Q^d\) is countable, but it is not Euclidean-open because \(\mathbb Q^d\) is dense.
<1>4. These observations also distinguish all six topologies. In particular, the cofinite topology is strictly below both the Euclidean and cocountable topologies, while the discrete topology is strictly above all of them.
Separable space (Tag E-ARJT3)
- What is a separable space?
Solution
<1>1. Definition of a separable space: <2>1. A topological space \((X, \mathcal{T})\) is called separable if there exists a subset \(D \subseteq X\) such that: 1. \(D\) is at most countable (i.e., \(|D| \le \aleph_0\)), 2. \(D\) is dense in \(X\), meaning that the topological closure of \(D\) equals \(X\): \[ \overline{D} = X. \]
<1>2. Equivalent characterizations: <2>1. A space \(X\) is separable if and only if there exists a countable set \(D \subseteq X\) that intersects every non-empty open set: \[ \forall U \in \mathcal{T} \setminus \{\emptyset\}, \qquad U \cap D \neq \emptyset. \]
<1>3. Properties and examples: <2>1. Every second-countable space is separable (choosing one point from each element of a countable basis). <2>2. For metric spaces, separability is equivalent to being second-countable and equivalent to being Lindelöf. <2>3. The Euclidean space \(\mathbb{R}^n\) with the standard topology is separable, with countable dense subset \(\mathbb{Q}^n\).
<1>4. Conclusion: A separable space is a topological space containing a countable dense subset. Q.E.D.
Nowhere dense subspace (Tag E-45N4G)
- What is a nowhere dense subspace?
Solution
<1>1. A subset \(A\subseteq X\) is nowhere dense in \(X\) if \[ \operatorname{int}_X(\overline A)=\varnothing. \]
<1>2. Equivalently, \[ X\setminus\overline A \] is dense in \(X\). Indeed, \[ \overline{X\setminus\overline A} =X\setminus\operatorname{int}(\overline A). \]
<1>3. Equivalently, every nonempty open set \(U\subseteq X\) contains a nonempty open set \(V\subseteq U\) with \(V\cap A=\varnothing\). <2>1. If \(A\) is nowhere dense, take \[ V=U\cap(X\setminus\overline A), \] which is nonempty because \(X\setminus\overline A\) is dense. <2>2. Conversely, if every nonempty open \(U\) contains such a \(V\), then no nonempty open set can be contained in \(\overline A\); hence \(\operatorname{int}(\overline A)=\varnothing\).
<1>4. For example, finite subsets of \(\mathbb R\), the set \(\mathbb Z\), and the Cantor set are nowhere dense in \(\mathbb R\). The rationals are not: \(\overline{\mathbb Q}=\mathbb R\).
$\QQ$ (Tag P-MFWBK)
- \(\QQ\)
Solution
No operation or property to determine is specified for \(\mathbb Q\).
$\ZZ$ (Tag P-HPN6K)
\(\ZZ\)
\(\ts{1}\)
\(\ts{p \in \ZZ^{\geq 0} \st p\text{ is prime}}\)
\(\ts{ {1\over n} \st n\in \ZZ^{\geq 0}}\)
\(\ts{ {1\over n} \st n\in \ZZ^{\geq 0}} \union \ts{0}\)
Prove that \(\RR^n\) is not homeomorphic to \(\RR\) for any \(n\geq 2\).
Is it true that the closure of a product is the product of the closures?
Solution
<1>1. No operation or property to determine is specified for the first five displayed subsets.
<1>2. For every \(n\ge2\), \(\mathbb R^n\) is not homeomorphic to \(\mathbb R\).
If \(h:\mathbb R^n\to\mathbb R\) were a homeomorphism, then deleting any point \(x\) would give a homeomorphism \(\mathbb R^n\setminus\{x\}\cong\mathbb R\setminus\{h(x)\}\). For \(n\ge2\), \(\mathbb R^n\setminus\{x\}\) is path-connected, while \(\mathbb R\setminus\{h(x)\}\) has two connected components, a contradiction.
<1>3. For arbitrary subsets \(A\subseteq X\) and \(B\subseteq Y\), \[\boxed{\overline{A\times B}^{\,X\times Y}=\overline A^{\,X}\times\overline B^{\,Y}.}\]
If \((x,y)\) lies in the left closure, continuity of the projections gives \(x\in\overline A\) and \(y\in\overline B\). Conversely, if \(x\in\overline A\) and \(y\in\overline B\), every basic neighborhood \(U\times V\) of \((x,y)\) meets \(A\times B\) because \(U\cap A\ne\varnothing\) and \(V\cap B\ne\varnothing\).
Interior of a product versus product of interiors (Tag P-N7RR5)
Is it true that the interior of a product is the product of the interiors? Specifically, for topological spaces \(X\) and \(Y\) and subsets \(A \subseteq X\) and \(B \subseteq Y\), is \(\operatorname{int}(A \times B) = \operatorname{int}(A) \times \operatorname{int}(B)\)? What about infinite products?
Solution
<1>1. For two factors, \[ \operatorname{int}_{X\times Y}(A\times B) =\operatorname{int}_X(A)\times\operatorname{int}_Y(B). \] <2>1. The right-hand side is open in \(X\times Y\) and contained in \(A\times B\), so it is contained in the left-hand side. <2>2. Conversely, if \((x,y)\in\operatorname{int}(A\times B)\), then some basic open set \(U\times V\) satisfies \[ (x,y)\in U\times V\subseteq A\times B. \] Hence \(U\subseteq A\) and \(V\subseteq B\), so \(x\in\operatorname{int}(A)\) and \(y\in\operatorname{int}(B)\).
<1>2. The same argument gives the finite-product identity \[ \operatorname{int}\!\left(\prod_{i=1}^n A_i\right) =\prod_{i=1}^n\operatorname{int}(A_i). \]
<1>3. Let now \(\prod_{i\in I}X_i\) have the product topology and put \(A=\prod_{i\in I}A_i\). <2>1. If \(A\) has nonempty interior, choose a basic open set \[ \prod_{i\in I}U_i\subseteq A, \] where \(U_i=X_i\) for all but finitely many \(i\). <2>2. For every index with \(U_i=X_i\), the inclusion \(U_i\subseteq A_i\) forces \(A_i=X_i\). Thus \(A_i=X_i\) for all but finitely many \(i\). <2>3. Conversely, if \(A_i=X_i\) for all but finitely many \(i\), then \[ \operatorname{int}(A)=\prod_{i\in I}\operatorname{int}(A_i), \] because the right-hand side is then a basic open product (possibly empty), and the finite-coordinate argument from <1>1 proves maximality. <2>4. Therefore \[ \operatorname{int}\!\left(\prod_{i\in I}A_i\right)= \begin{cases} \prod_{i\in I}\operatorname{int}(A_i),&A_i=X_i\text{ for all but finitely many }i,\\ \varnothing,&\text{otherwise}. \end{cases} \]
<1>4. Hence the finite-product identity always holds, whereas for infinite products it can fail. For example, \[ \operatorname{int}_{\mathbb R^{\mathbb N}}([0,1]^{\mathbb N})=\varnothing \quad\text{but}\quad \prod_{n\ge1}\operatorname{int}[0,1]=(0,1)^{\mathbb N}\ne\varnothing. \]
Connected versus locally connected spaces (Tag P-2V6GL)
- Find a space that is connected but not locally connected. Can there be a space that is locally connected but not connected?
Solution
Goal. Find a connected but not locally connected space, and answer the converse question.
<1>1. A connected but not locally connected space: the topologist’s sine curve. <2>1. \(X = \theset{(x, \sin(1/x)) : x > 0} \cup \theset{(0, y) : -1 \le y \le 1}\).
the topologist’s sine curve.
<2>2. \(X\) is connected.
the graph \(\theset{(x, \sin(1/x)) : x > 0}\) is connected (continuous image of \((0,\infty)\)), and its closure is \(X\), so \(X\) is connected.
<2>3. \(X\) is not locally connected at \(p=(0,0)\).
Let \[ U=X\cap\bigl(\mathbb R\times(-1/2,1/2)\bigr), \] an open neighborhood of \(p\) in \(X\). We claim that the connected component of \(p\) in \(U\) is exactly \[ C=\{0\}\times(-1/2,1/2). \] Certainly \(C\) is connected and contains \(p\).
Now let \[ q=(x_q,\sin(1/x_q))\in U \] be any point of the oscillating graph, so \(x_q>0\). Choose \(m\) sufficiently large that \[ c=\frac1{\pi/2+2\pi m}<x_q. \] Then \(\sin(1/c)=1\), so \(U\) contains no point whose first coordinate is \(c\). Consequently \[ U_- = U\cap\{x<c\},\qquad U_+=U\cap\{x>c\} \] form a separation of \(U\): both are relatively open, and because \(U\) has no point with first coordinate \(c\), each is also the complement of the other in \(U\). We have \(p\in U_-\) and \(q\in U_+\). Hence no connected subset of \(U\) containing \(p\) can contain any graph point. This proves that the component of \(p\) in \(U\) is precisely \(C\).
Finally, \(C\) is not a neighborhood of \(p\) in \(X\). Every Euclidean neighborhood of \((0,0)\) meets the graph \(\{(x,\sin(1/x)):x>0\}\) at points with \(x>0\) and \(\sin(1/x)\) arbitrarily close to \(0\). Thus there is no connected open set \(W\) with \[ p\in W\subseteq U, \] because any such \(W\) would have to lie in the component \(C\) but \(C\) contains no neighborhood of \(p\). Therefore \(X\) is not locally connected at \(p\).
<1>2. A locally connected but not connected space exists. <2>1. Example: the disjoint union of two intervals, \(X = (0,1) \cup (2,3)\).
a disjoint union of two open intervals.
<2>2. \(X\) is locally connected.
each point has a connected neighborhood (a small interval around it).
<2>3. \(X\) is not connected.
it is a disjoint union of two nonempty open sets.
<1>3. Q.E.D.
<1>1 gives a connected but not locally connected space; <1>2 gives a locally connected but not connected space (yes, such spaces exist).
One-point compactification is compact; path-connected implies connected (Tag P-3GOJI)
Show that for \(X\) an arbitrary topological space, the one-point (Alexandroff) compactification \(X^* = X \cup \{\infty\}\) is compact.
Prove that every path-connected topological space is connected.
Solution
<1>1. Let \(X^*=X\cup\{\infty\}\) carry the Alexandroff topology: the open sets not containing \(\infty\) are the open subsets of \(X\), while a neighborhood of \(\infty\) has the form \[ \{\infty\}\cup(X\setminus K) \] with \(K\subseteq X\) compact and closed.
<1>2. The space \(X^*\) is compact.
Let \(\mathcal U\) be an open cover of \(X^*\). Choose \(U_\infty\in\mathcal U\) containing \(\infty\). Then for some compact closed \(K\subseteq X\), \[ \{\infty\}\cup(X\setminus K)\subseteq U_\infty. \] The remaining points lie in \(K\). The sets \(U\cap X\), for \(U\in\mathcal U\), cover \(K\), so compactness of \(K\) gives finitely many of them covering \(K\). Together with \(U_\infty\) they form a finite subcover of \(X^*\).
<1>3. Every path-connected space is connected.
Fix \(x_0\in X\). For each \(x\in X\), choose a path from \(x_0\) to \(x\). Each path image is connected, all such images contain \(x_0\), and their union is \(X\). A union of connected subsets with a common point is connected.
The topologist's sine curve is connected but not path-connected (Tag P-JAEYU)
- Show that the topologist’s sine curve is connected but not path-connected.
Solution
<1>1. Let \[G=\{(x,\sin(1/x)):0<x\le1\},\qquad T=G\cup(\{0\}\times[-1,1]).\] The set \(T\) is connected.
The interval \((0,1]\) is connected, so its continuous image \(G\) is connected. Its closure is exactly \(T\), and the closure of a connected subset is connected.
<1>2. There is no path in \(T\) from the vertical segment \(\{0\}\times[-1,1]\) to \(G\).
Suppose \(\gamma(t)=(x(t),y(t))\) were such a path, with \(x(0)=0\) and \(x(1)>0\). Let \[t_0=\sup\{t\in[0,1]:x(t)=0\}.\] Then \(t_0<1\), \(x(t_0)=0\), and \(x(t)>0\) for \(t>t_0\). For any \(t_1>t_0\), continuity of \(x\) implies its image on \([t_0,t_1]\) contains every value between \(0\) and \(x(t_1)\). Hence for all sufficiently large \(n\) there are times tending to \(t_0\) at which \[x=\frac1{\pi/2+2\pi n}\quad\text{and}\quad x=\frac1{3\pi/2+2\pi n}.\] At these times \(y=1\) and \(y=-1\), respectively. The times must tend to \(t_0\): on every compact interval \([t_0+\delta,t_1]\), the positive continuous function \(x\) has a positive minimum. Thus \(y(t)\) cannot converge to the single value \(y(t_0)\), contradicting continuity of \(\gamma\).
<1>3. Therefore the topologist’s sine curve is connected but not path-connected.
Use <1>1 and <1>2.
Products of Hausdorff spaces; $\RR$ versus $[0,\infty)$ (Tag P-FNJCM)
Is every product (finite or infinite) of Hausdorff spaces Hausdorff?
Is \(\RR\) homeomorphic to \([0, \infty)\)?
Solution
<1>1. Every product, finite or infinite, of Hausdorff spaces is Hausdorff.
If two points \(x=(x_i)\) and \(y=(y_i)\) differ, choose a coordinate \(j\) with \(x_j\ne y_j\). Separate \(x_j,y_j\) by disjoint open sets \(U,V\) in the Hausdorff factor \(X_j\). Their inverse images under the projection \(\prod_iX_i\to X_j\) are disjoint open neighborhoods of \(x,y\).
<1>2. The spaces \(\mathbb R\) and \([0,\infty)\) are not homeomorphic.
Removing any point from \(\mathbb R\) disconnects it into two components. But removing the endpoint \(0\) from \([0,\infty)\) leaves \((0,\infty)\), which is connected. A homeomorphism would preserve this point-deletion property.
A space is connected iff its only clopen subsets are $\emptyset$ and $X$ (Tag P-WXTVX)
Let \(X\) be a topological space. Prove that \(X\) is connected if and only if the only subsets of \(X\) that are both open and closed (clopen) are \(\varnothing\) and \(X\).
Solution
<1>1. Suppose \(X\) is connected and \(A\subseteq X\) is clopen. If \(A\) were neither \(\varnothing\) nor \(X\), then \[ X=A\sqcup(X\setminus A) \] would be a union of two disjoint nonempty open sets, contradicting connectedness. Hence the only clopen subsets are \(\varnothing\) and \(X\).
<1>2. Conversely, suppose the only clopen subsets are \(\varnothing\) and \(X\). If \(X\) were disconnected, there would be disjoint nonempty open sets \(U,V\) with \(X=U\cup V\). Then \[ X\setminus U=V \] is open, so \(U\) is also closed. Thus \(U\) is a nontrivial clopen subset, a contradiction.
<1>3. Therefore \(X\) is connected if and only if its only clopen subsets are \(\varnothing\) and \(X\).
Closed subsets of compact spaces are compact (Tag P-NNHWB)
- Show that a closed subset \(A\) of a compact space \(X\) is compact.
Solution
<1>1. Let \(\mathcal U\) be an open cover of the closed subset \(A\subseteq X\).
We prove that \(\mathcal U\) has a finite subcover.
<1>2. Since \(A\) is closed, \(X\setminus A\) is open, and \[\mathcal U\cup\{X\setminus A\}\] is an open cover of \(X\).
The members of \(\mathcal U\) cover \(A\), while \(X\setminus A\) covers every point outside \(A\).
<1>3. Compactness of \(X\) gives a finite subcover of this enlarged cover.
This is the definition of compactness.
<1>4. Removing \(X\setminus A\) if it occurs leaves finitely many members of \(\mathcal U\) that cover \(A\). Hence \(A\) is compact.
The removed set contains no point of \(A\).
Open subsets of compact spaces (Tag P-OTXNQ)
Does this hold when \(A\) is instead an open subset?
Solution
The preceding assertion is needed to determine which property is being asked about for open subsets.
Continuous images of compact spaces, and compact metric spaces are complete (Tag P-NJNNL)
- Let \(X\) and \(Y\) be topological spaces, and let \(f: X \to Y\) be a continuous map. Prove that if \(X\) is compact, then its image \(f(X) \subseteq Y\) is compact.
- Let \((X, d)\) be a metric space. Prove that if \(X\) is compact, then \((X, d)\) is complete (every Cauchy sequence in \(X\) converges to a limit in \(X\)).
Solution
<1>1. If \(X\) is compact and \(f:X\to Y\) is continuous, then \(f(X)\) is compact.
Let \(\{V_\alpha\}\) be an open cover of \(f(X)\). Then \(\{f^{-1}(V_\alpha)\}\) is an open cover of \(X\). Compactness gives finitely many indices \(\alpha_1,\dots,\alpha_r\) with \[ X=\bigcup_{j=1}^r f^{-1}(V_{\alpha_j}). \] Applying \(f\) shows \[ f(X)\subseteq\bigcup_{j=1}^r V_{\alpha_j}, \] so the original cover has a finite subcover.
<1>2. Every compact metric space is complete.
Let \((x_n)\) be Cauchy in compact metric \(X\). Compact metric spaces are sequentially compact, so some subsequence \(x_{n_k}\) converges to a point \(x\in X\).
Given \(\varepsilon>0\), choose \(N\) so that \[ d(x_m,x_n)<\varepsilon/2\qquad(m,n\ge N). \] Choose \(k\) with \(n_k\ge N\) and \(d(x_{n_k},x)<\varepsilon/2\). Then for every \(n\ge N\), \[ d(x_n,x)\le d(x_n,x_{n_k})+d(x_{n_k},x)<\varepsilon. \] Thus \(x_n\to x\), so every Cauchy sequence converges in \(X\).
Compact subsets of Hausdorff spaces are closed (Tag P-SG462)
Show that a compact subset of a Hausdorff space is closed.
Solution
<1>1. Let \(A\) be a compact subset of a Hausdorff space \(X\). To prove that \(A\) is closed, it suffices to prove that \(X\setminus A\) is open.
A subset of a topological space is closed exactly when its complement is open.
<1>2. Fix \(x\in X\setminus A\). For every \(a\in A\), choose disjoint open sets \[ U_a\ni a, \qquad V_a\ni x. \]
Since \(x\notin A\), one has \(x\ne a\) for every \(a\in A\). Because \(X\) is Hausdorff, the distinct points \(a\) and \(x\) admit disjoint open neighborhoods \(U_a\) and \(V_a\).
<1>3. There are finitely many points \(a_1,\dots,a_m\in A\) such that \[ A\subseteq U_{a_1}\cup\cdots\cup U_{a_m}. \]
The family \[ \{U_a:a\in A\} \] is an open cover of the compact set \(A\). Compactness therefore gives a finite subcover.
<1>4. The set \[ V=V_{a_1}\cap\cdots\cap V_{a_m} \] is an open neighborhood of \(x\) disjoint from \(A\).
It is open as a finite intersection of open sets, and it contains \(x\) because every \(V_{a_i}\) contains \(x\).
Suppose \(y\in V\cap A\). By <1>3, \(y\in U_{a_i}\) for some \(i\). But \(y\in V\subseteq V_{a_i}\) as well, contradicting \[ U_{a_i}\cap V_{a_i}=\varnothing. \] Thus \(V\cap A=\varnothing\).
<1>5. Therefore \(A\) is closed in \(X\).
By <1>4, every point \(x\in X\setminus A\) has an open neighborhood contained in \(X\setminus A\). Hence \(X\setminus A\) is open. By <1>1, \(A\) is closed.
Closed versus compact subsets of Hausdorff spaces (Tag P-GAA3C)
Does the converse hold?
Solution
The preceding assertion is needed to determine which converse is being asked about.
When compact sets are closed, and $\overline{A\times B}=\overline{A}\times\overline{B}$ (Tag P-YBQ3V)
What property on a space guarantees that compact sets are closed
What property on a space guarantees that closed sets are compact?
Show that a continuous bijection from a compact space to a Hausdorff space is necessarily a homeomorphism.
Solution
<1>1. In a Hausdorff space every compact subset is closed.
For a compact \(K\) and \(x\notin K\), separate \(x\) from each point of \(K\) and use compactness to obtain one neighborhood of \(x\) disjoint from \(K\).
<1>2. In a compact space every closed subset is compact.
If \(F\) is closed and \(\mathcal U\) covers \(F\), then \(\mathcal U\cup\{X\setminus F\}\) covers \(X\); a finite subcover restricts to one of \(F\).
<1>3. A continuous bijection \(f:X\to Y\) from compact \(X\) to Hausdorff \(Y\) is a homeomorphism.
For closed \(F\subseteq X\), \(F\) is compact, so \(f(F)\) is compact and therefore closed in \(Y\). Thus \(f\) is a closed bijection, hence \(f^{-1}\) is continuous.
<1>4. For arbitrary subsets \(A\subseteq X\) and \(B\subseteq Y\), \[\boxed{\overline{A\times B}^{\,X\times Y}=\overline A^{\,X}\times\overline B^{\,Y}.}\]
If \((x,y)\) is in the left closure, projections show \(x\in\overline A\) and \(y\in\overline B\). Conversely, if \(x\in\overline A\) and \(y\in\overline B\), every basic neighborhood \(U\times V\) of \((x,y)\) meets \(A\times B\) because \(U\cap A\) and \(V\cap B\) are both nonempty.
Closed sets covering a connected space with connected intersection are connected (Tag P-ZQBPZ)
Let \(X\) be a connected space and \(A,B\subseteq X\) closed subsets with \(X=A\cup B\) and \(A\cap B\) connected. Show that \(A\) and \(B\) are connected.
Solution
Goal. For connected \(X = A \cup B\) with \(A, B\) closed and \(A \cap B\) connected, show \(A\) and \(B\) are connected.
<1>1. Suppose \(A\) is disconnected.
assume for contradiction.
<1>2. Then \(A = C \cup D\) with \(C, D\) nonempty, disjoint, and closed in \(A\) (hence closed in \(X\), since \(A\) is closed).
a disconnected space is a union of two nonempty disjoint closed subsets.
<1>3. \(A \cap B\) is connected, so it lies entirely in \(C\) or entirely in \(D\).
\(A \cap B = (C \cap B) \cup (D \cap B)\) is a union of two disjoint closed sets; since \(A \cap B\) is connected, one of them is empty, so \(A \cap B \subseteq C\) or \(A \cap B \subseteq D\).
<1>4. WLOG \(A \cap B \subseteq C\).
relabel if necessary.
<1>5. Then \(X = (C \cup B) \cup D\) is a union of two disjoint nonempty closed sets. <2>1. \(C \cup B\) and \(D\) are disjoint.
\(C \cap D = \emptyset\) and \(B \cap D = \emptyset\) (since \(A \cap B \subseteq C\) and \(D \subseteq A\)). <2>2. \(C \cup B\) and \(D\) are closed.
\(C, D\) are closed in \(X\), and \(B\) is closed, so \(C \cup B\) is closed.
<2>3. Both are nonempty.
\(C \neq \emptyset\) and \(D \neq \emptyset\).
<2>4. \(X = (C \cup B) \cup D\).
\(X = A \cup B = (C \cup D) \cup B = (C \cup B) \cup D\).
<1>6. This contradicts \(X\) being connected.
\(X\) is a union of two disjoint nonempty closed sets.
<1>7. Hence \(A\) is connected; by symmetry, \(B\) is connected.
the same argument with \(A\) and \(B\) swapped.
<1>8. Q.E.D.
<1>7 is the claim.
Strict implications $T_4\Rightarrow T_3\Rightarrow T_2\Rightarrow T_1\Rightarrow T_0$ (Tag P-WHXWM)
- Prove the following implications of separation axioms, and show that
they are strict:
Solution
Goal. Prove the implications \(T_4 \Rightarrow T_3 \Rightarrow T_2 \Rightarrow T_1 \Rightarrow T_0\) and show each is strict.
<1>1. \(T_4 \Rightarrow T_3\). <2>1. \(T_4\) means normal and \(T_1\); \(T_3\) means regular and \(T_1\).
definitions.
<2>2. A normal \(T_1\) space is regular.
given a closed set \(C\) and a point \(x \notin C\), since the space is \(T_1\) the singleton \(\theset{x}\) is closed, and normality separates the disjoint closed sets \(C\) and \(\theset{x}\) by disjoint open sets.
<1>2. \(T_3 \Rightarrow T_2\). <2>1. A regular \(T_1\) space is Hausdorff.
given distinct \(x, y\), \(T_1\) makes \(\theset{y}\) closed and \(x \notin \theset{y}\); regularity separates \(x\) and \(\theset{y}\) by disjoint open sets.
<1>3. \(T_2 \Rightarrow T_1\). <2>1. A Hausdorff space is \(T_1\).
given distinct \(x, y\), Hausdorff gives disjoint open \(U \ni x\), \(V \ni y\); then \(x \notin V\) and \(y \notin U\), so each singleton is closed.
<1>4. \(T_1 \Rightarrow T_0\). <2>1. A \(T_1\) space is \(T_0\).
given distinct \(x, y\), \(T_1\) gives an open set containing \(x\) but not \(y\) (namely \(X \sm \theset{y}\)).
<1>5. Each implication is strict. <2>1. \(T_0\) but not \(T_1\): the Sierpiński space \(\theset{0,1}\) with open sets \(\theset{\emptyset, \theset{1}, \theset{0,1}}\).
\(1\) has a neighborhood not containing \(0\), but \(\theset{0}\) is not closed.
<2>2. \(T_1\) but not \(T_2\): the cofinite topology on an infinite set.
singletons are closed, but any two nonempty open sets intersect.
<2>3. \(T_2\) but not \(T_3\): let \(K=\{1/n:n\ge1\}\) and give \(\mathbb R\) the \(K\)-topology generated by the basis \[ \{(a,b):a<b\}\;\cup\;\{(a,b)\setminus K:a<b\}. \] This space is Hausdorff but not regular.
The \(K\)-topology contains the usual topology, so it is Hausdorff. Also \(K\) is closed: for every \(x\notin K\), some basis set of the form \((a,b)\setminus K\) contains \(x\) and misses \(K\).
Suppose regularity separated the point \(0\) from the closed set \(K\) by disjoint open sets \(U\ni0\) and \(V\supseteq K\). Shrinking \(U\) to a basic neighborhood, we may choose \(\varepsilon>0\) with \[ (-\varepsilon,\varepsilon)\setminus K\subseteq U. \] Choose \(1/n<\varepsilon/2\). Since \(1/n\in V\), a basic neighborhood of \(1/n\) is contained in \(V\). Any basic neighborhood containing \(1/n\) must contain an ordinary interval \((c,d)\) about \(1/n\), because sets of the form \((a,b)\setminus K\) omit \(1/n\). This interval contains a point \[ y\in(-\varepsilon,\varepsilon)\setminus K, \] so \(y\in U\cap V\), a contradiction. Thus the space is not regular.
<2>4. \(T_3\) but not \(T_4\): the Sorgenfrey plane \(S\times S\), where \(S\) is \(\mathbb R\) with basis \([a,b)\).
First, \(S\) has a basis of clopen sets: each \([a,b)\) is open, and its complement \[ (-\infty,a)\cup[b,\infty) \] is open in the Sorgenfrey topology. Hence \(S\) is regular and \(T_1\). Products of clopen basic sets form a clopen basis for \(S\times S\), so the Sorgenfrey plane is also regular and \(T_1\), hence \(T_3\).
It remains to prove that it is not normal. The countable set \(\mathbb Q\times\mathbb Q\) is dense, since every basic rectangle \([a,b)\times[c,d)\) contains a rational pair. On the other hand, the anti-diagonal \[ D=\{(x,-x):x\in\mathbb R\} \] is a closed discrete subspace of cardinality \(\mathfrak c\). It is discrete because \[ [x,x+\varepsilon)\times[-x,-x+\varepsilon) \] meets \(D\) only at \((x,-x)\). It is closed because, if \(a+b\ne0\), a sufficiently small basic rectangle about \((a,b)\) keeps the sign of the coordinate sum and therefore misses \(D\).
Now use the following Jones lemma. A separable normal \(T_1\) space cannot contain a closed discrete subspace of cardinality \(\mathfrak c\). Indeed, if \(D\) were such a subspace, then every subset \(E\subseteq D\) and its complement \(D\setminus E\) would be disjoint closed subsets of the ambient space. By Urysohn’s lemma there would be a continuous map \(f_E\) to \([0,1]\) taking value \(0\) on \(E\) and \(1\) on \(D\setminus E\). Distinct subsets give distinct continuous maps, so there would be at least \(2^{\mathfrak c}\) continuous maps to \([0,1]\). But on a separable space a continuous map to the Hausdorff space \([0,1]\) is determined by its values on a fixed countable dense subset, so there are at most \[ |[0,1]|^{\aleph_0}=\mathfrak c \] such maps, contradicting Cantor’s theorem \(2^{\mathfrak c}>\mathfrak c\).
Thus the separable Sorgenfrey plane is not normal, so it is \(T_3\) but not \(T_4\).
<1>6. Q.E.D.
<1>1–<1>4 prove the implications; <1>5 shows each is strict.
Every compact metrizable space has a countable basis (Tag P-TC2PJ)
- Show that every compact metrizable space has a countable basis.
Solution
<1>1. For each \(n \ge 1\), the open balls \(\{B(x, 1/n) : x \in X\}\) form an open cover of \(X\).
every point lies in its own ball.
<1>2. Since \(X\) is compact, there is a finite subcover \(\{B(x_{n,1}, 1/n), \ldots, B(x_{n,k_n}, 1/n)\}\).
compactness.
<1>3. Let \(\mathcal B = \{B(x_{n,j}, 1/n) : n \ge 1,\ 1 \le j \le k_n\}\).
collect all these balls.
<1>4. \(\mathcal B\) is countable.
it is a countable union of finite sets.
<1>5. \(\mathcal B\) is a basis for the topology of \(X\). <2>1. Let \(U\) be open and \(x \in U\).
take an arbitrary point of an arbitrary open set.
<2>2. There is \(\epsilon > 0\) with \(B(x, \epsilon) \subseteq U\).
\(U\) is open.
<2>3. Choose \(n\) with \(1/n < \epsilon/2\).
Archimedean property.
<2>4. Since \(\{B(x_{n,j}, 1/n)\}\) covers \(X\), there is \(j\) with \(x \in B(x_{n,j}, 1/n)\).
<1>2.
<2>5. Then \(B(x_{n,j}, 1/n) \subseteq B(x, \epsilon) \subseteq U\).
if \(y \in B(x_{n,j}, 1/n)\), then \(d(x,y) \le d(x, x_{n,j}) + d(x_{n,j}, y) < 1/n + 1/n = 2/n < \epsilon\).
<2>6. Hence every point of \(U\) lies in a member of \(\mathcal B\) contained in \(U\).
<2>1–<2>5.
<1>6. Therefore \(\mathcal B\) is a countable basis.
<1>4 and <1>5.
<1>7. Q.E.D.
<1>6.