The connected sum of $X$ and $Y$ is the surface$$X \# Y:=(X \backslash\mathring D) ⊔\left(Y \backslash\mathring{D'}\right) / ∼$$ where $D ⊂X$ and $D' ⊂Y$ are embedded discs, and the equivalence relation ∼ identifies $∂ D$ and $∂ D'$ via some homeomorphism $φ: ∂ D → ∂ D'$ [Not required: If $X$ and $Y$ are surfaces with boundary, then $D$ and $D'$ should not meet the boundary].
Every closed surface is homeomorphic to either $S^2$, a connected sum of copies of the torus $T^2$, or a connected sum of copies of the projective plane $ℝP^2$. Moreover, the surfaces in the above list are pairwise non-homeomorphic.
The Möbius band is homeomorphic to $ℝP^2$ with one disc removed: Lemma 5.12
To see this, consider the following construction of $ℝP^2$:
and remove this disc:
So the connected sum of two Möbius bands is homeomorphic to $ℝP^2 \# ℝP^2$ with two discs removed.
It remains to show that $ℝP^2 \# ℝP^2$ is homeomorphic to a Klein bottle: Example 5.22 The boundary words xy^{−1}xy and xxyy represent homeomorphic surfaces.
Yes: $S^1$ minus a point is homeomorphic to an open interval. The latter is homeomorphic to ℝ.
β) is it homeomorphic to a closed subspace of $S^1$ ?No: Any closed subset of $S^1$ is compact, but ℝ is not compact.
No: Any subset of ℝ (with at least three points) has the property that it becomes disconnected after removing some point. But $S^1$ is connected and remains connected after removing any point.
If $U$ is open, then $U ∩[a, b]$ is open in $[a, b]$ by definition of the subspace topology. Assume that $U ∩[a, b]$ is open in $[a, b]$ for all $a, b∈ ℝ$. We need to show that for every point $x∈ U$ there exists $ε>0$ such that $(x-ε, x+ε) ⊂U$. Pick any interval $[a, b]$ containing $x$ in its interior. Since $U ∩[a, b]$ is open in $[a, b]$, there exists $ε>0$ such that $(x-ε, x+ε) ⊂U ∩[a, b]$. In particular, $(x-ε, x+ε) ⊂U$.
(If $U$ is open, then $U ∩ V$ is open in $V$ by definition of the subspace topology.) Assume $U ∩ V$ is open in $V$ for all closed subsets $V ⊊ X$. We'll show that $U$ is open by proving that every point $x∈ U$ admits an open neighbourhood contained in $U$. Pick $y∈ X$ distinct from $x$ (if $X$ has only one point, then $U$ is trivially open). Since $X$ is Hausdorff, there exist disjoint open subsets $A, B ⊂X$ with $x∈ A$ and $y∈ B$. Let $V:=X \backslash B$, so that$$x∈ A ⊂V ⊊ X$$By assumption, $U ∩ V$ is open in $V$. Therefore $U ∩ A$ is open in $A$. Since $A$ is open in $X, U ∩ A$ is also open in $X$. $U ∩ A$ is the desired neighbourhood of $x$.
β) Prove that the answer is "no" in general.Let $X$ be the two point space equipped with the indiscrete topology, and let $U$ be a subspace consisting of just one of the two points. $U$ is not open, but $U ∩ ∅$ is open in $∅$ (and $∅$ is the only closed subset of $X$ which is not the whole space).
A topological space $X$ is compact if every open cover $\left\{U_i\right\}_{i∈ I}$ (an open cover is a collection of open sets whose union is $X$) admits a subcover $\left\{U_i\right\}_{i∈ J}$ (a subcover is a sub-collection whose union is still $X$) where $J ⊂I$ is finite.
Given $x∈ ℝ$, we have $x∈(x-1, x+1) ⊂[x-1, x+1]$. In other words, $[x-1, x+1]$ is a compact neighbourhood of $x$.
β) Is ℚ (equipped with the subspace topology from ℝ) locally compact? Justify.No point $x∈ℚ$ admits a compact neighbourhood. If $V ⊂ℚ$ is a neighbourhood of $x$, then $[x-ε, x+ε]∩ℚ⊂V$ for some $ε>0$. Since $[x-ε, x+ε]∩ℚ$ is not complete, it is not compact, but $[x-ε, x+ε]∩ℚ$ is closed in $V$, so $V$ is not compact.
We may assume without loss of generality (by performing a translation) that $V$ contains 0 in its interior. Let $ρ_θ∈ ℝ⩾ 0$ be defined by the requirement that $\left[0, ρ_θ ⋅ θ\right]=V ∩\left(ℝ_{⩾ 0} ⋅ θ\right)$. Note that this is well-defined: since $V$ is convex, the intersection $V ∩\left(ℝ_{⩾ 0} ⋅ θ\right)$ is also convex, in particular connected. The function $θ ↦ ρ_θ$ is clearly continuous (it is piecewise algebraic). The map $D^2 → V$ given, in polar coordinates, by $(r, θ) ↦\left(ρ_θ r, θ\right)$ is the desired homeomorphism. It is continuous at non-zero points because $ρ_θ$ is continuous. It is continuous at zero because $ρ_θ$ is bounded. Finally, it maps bijectively $D^2$ onto $V$, and is thus a homeomorphism (a continuous bijection between compact Hausdorff spaces is a homeomorphism).
Let $T ⊂ℝ^2$ be the triangle with vertices $(0,0),(1,0)$ and $(1,1)$. The map $f:[0,1]^2 →T$ given by $f(x, y)=(x, x y)$ is continuous and surjective. It satisfies $f(x, y)=f\left(x', y'\right)$ iff $(x, y) ∼\left(x', y'\right)$ therefore it descends to a continuous map (indeed a continuous bijection) $\bar{f}:[0,1]^2 /\mmlToken{mi}∼ → T$. The $\operatorname{map} \bar{f}$ is a homeomorphism (because a continuous bijection from a compact space to a Hausdorff space is a homeomorphism: $[0,1]^2/\mmlToken{mi}∼$ is compact because it's the image of $[0,1]^2$ under a continuous map; $T$ is Hausdorff because it's a subspace of $ℝ^2$ ). Finally, $T$ is homeomorphic to $D^2$ by (b.i).
$X_1:=D^2 \backslash\{(1,0)\}$ | $X_2:=\stackrel{∘}{D}^2 ∪\left(D^2 ∩ H\right)$ | $X_3:=\stackrel{∘}{D}^2 ∪\left(D^2 ∩ \stackrel{∘}{H}\right)$ | $X_4:=\stackrel{∘}{D}^2 ∪\{(1,0)\}$ |
Let $\bar{f}:[0,1]^2 /\mmlToken{mi}∼ → T$ be the homeomorphism constructed in part (b), and let $p∈[0,1]^2 /\mmlToken{mi}∼$ be the point corresponding to the subset $A:=\{0\} ×[0,1] ⊂[0,1]^2$ (note that $A$ is an equivalence class for the equivalence relation ∼). By (b.i), we have the following two homeomorphisms: $X_1≅T \backslash\{(0,0)\}$ and $X_3≅[0,1]^2 \backslash A$. The map $\bar{f}$ induces a homeomorphism $\left([0,1]^2 /\mmlToken{mi}∼\right) \backslash\{p\}≅T \backslash\{(0,0)\}$. The composite map $[0,1]^2 \backslash A ↪[0,1]^2↠[0,1]^2 /\mmlToken{mi}∼$ induces a homeomorphism $[0,1]^2 \backslash A≅([0,1]^2 /\mmlToken{mi}∼)\backslash\{p\}$. Assembling all these maps, we get a homeomorphism$$X_3 \stackrel{≅}{⟶}[0,1]^2 \backslash A \stackrel{≅}{⟶}\left([0,1]^2 /\mmlToken{mi}∼\right) \backslash\{p\} \stackrel{≅}{⟶} T \backslash\{(0,0)\} \stackrel{≅}{⟶} X_1 .$$
We count the number of points that do not admit compact neighbourhoods: $X_1$ and $X_3$ admit no such points (they are locally compact). $X_2$ has two such points. $X_4$ has one such point.
Take $ℤ$ as the vertex set, and $\{\{n\}: n∈ ℤ\} ∪\{\{n, n+1\}: n∈ ℤ\}$ as the set of simplices.
The equivalence relation ∼ is given by declaring $x ∼ y$ when $f(x)=f(y)$.
β) Describe the unique topology on $Q$ which makes $\bar{f}$ into a homeomorphism (where $X / ∼$ is equipped with the quotient topology).A subspace $U ⊂Q$ is open in this topology iff $f^{-1}(U)$ is open in $X$.
$f$ is a quotient map if $\left\{\begin{aligned}&f\text{ is surjective}\\&U\text{ is open in $Y$ iff $f^{-1}(U)$ is open in $X$}\end{aligned}\right.$
Declare $x≡ x'$ if their images in $Z$ are equal. The composite $X → Y → Z$ induces a continuous bijection$$f: X /\mmlToken{mi}≡ → Z$$that fits into a commutative triangle Error! Click to view log. A subspace $U ⊂Z$ is open iff its preimage in $Y$ is open (by definition of the quotient topology on $Y /\mmlToken{mi}≈$) iff its preimage in $X$ is open (by definition of the quotient topology on $X /\mmlToken{mi}∼)$. Similarly, a subspace $U' ⊂X /\mmlToken{mi}≡$ is open iff its preimage in $X$ is open. The bijective map $f$ induces a bijection between the open subsets of $X /\mmlToken{mi}≡$ and the open subsets of $Z$. So $f$ is a homeomorphism.
The first space $S^2 / ∼$ is the standard definition of $ℝP^2$. The second space can be viewed as an iterated quotient$$\left(S^2 /(x, y, z) ∼(x, y,-z)\right) /(x, y, 0) ∼(-x,-y, 0)$$as in (c). The first quotient $S^2 /(x, y, z) ∼(x, y,-z)$ is homeomorphic to $D^2$ via the quotient map $S^2 → D^2:(x, y, z)↦(x, y)$ (a bijective continuous map between compact Hausdorff spaces is a homeomorphism). The equivalence relation $(x, y, 0) ∼(-x,-y, 0)$ on points of $S^2$ induces the equivalence relation $≈$ on $D^2$ that identifies antipodal points of the boundary. The quotient of $D^2 / ≈$ is another standard definition of $ℝP^2$. Explicitly, the embedding $D^2↪S^2:(x, y) ↦\left(x, y, \sqrt{1-x^2-y^2}\right)$ induces a homeomorphism $S^2 / ∼ → D^2 / ≈$ (a bijective continuous map between compact Hausdorff spaces is a homeomorphism).
$S^2 /(x, y, z) ∼(-x,-y,z)$ is isomorphic to the Klein bottle $N_2$.