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Let $a,b\in {\mathbb{R}}^2\setminus T$. If $a=b$, use the constant path. Otherwise consider the countably many lines through $a$ and a point of $T$, and through $b$ and a point of $T$. Choose a line $L$ whose direction is not parallel to any of these lines; such a direction exists because there are uncountably many directions but only a countable set of forbidden ones. Every forbidden line meets $L$ in at most one point. Choose $c\in L$ outside those countably many intersections and outside $\{a,b\}$.

Neither segment $[a,c]$ nor $[c,b]$ meets $T$: a forbidden point on either segment would put $c$ on the corresponding forbidden line. The piecewise linear pathlies in the complement. This proves that the plane minus a countable set is path connected.

If a homeomorphism $h:{\mathbb{R}}^2\to \mathbb{R}$ existed, its restriction after removing a point $a$ would give a homeomorphism between ${\mathbb{R}}^2\setminus \{a\}$ and $\mathbb{R}\setminus \{h(a)\}$. The first is a path-connected space and therefore connected; the second consists of two separated intervals. Since homeomorphisms preserve connectedness, $\boxed{{\mathbb{R}}^2\ncong \mathbb{R}}$.

The conclusions of (b) and (c) remain true. Indeed, the unit circle is a path-connected space, and the upper semicircle supplies a path from $-1$ to $1$; restricting the proposed closed cover or two-valued continuous function to that path reduces either claim to the corresponding result on $I$.

The fixed-point conclusion (d) fails. For example, any nontrivial spatial rotation $f(z)=e^{i\theta }z$ of the unit circle has no fixed point when $e^{i\theta }\ne 1$; the antipodal map $f(z)=-z$ is the simplest example.

A topological space is connected when it is not the union of two disjoint nonempty open subsets. It is path-connected when every two points $x,y$ can be joined by a path, meaning a continuous map $\gamma :[0,1]\to X$ with $\gamma (0)=x$ and $\gamma (1)=y$.

To prove that $[0,1]$ is connected, suppose it had a separation $[0,1]=U\cup V$ with $0\in U$. LetWhichever of $U$ or $V$ contains $c$, its relative openness supplies an interval about $c$ that contradicts either the definition of the supremum or the fact that all points immediately below $c$ lie in $U$. Thus no separation exists.

If $X$ is path-connected and $X=U\cup V$ were a separation, choose $u\in U$ and $v\in V$. A path from $u$ to $v$ would make $[0,1]$ the union of the disjoint nonempty open sets ${\gamma }^{-1}(U)$ and ${\gamma }^{-1}(V)$, contradicting the connectedness just proved. Hence

Now let $X\subseteq {\mathbb{R}}^n$ be open. Every point $x\in X$ lies in an open ball contained in $X$, and an open ball is path-connected by straight line segments. It follows that every path component of $X$ is open. If there were more than one path component, one component and the union of all the others would separate $X$. Consequently a connected open subset of Euclidean space is path-connected. The converse follows from the preceding implication.

The same argument answers the locally Euclidean question affirmatively. Every point has a neighbourhood homeomorphic to an open subset of ${\mathbb{R}}^n$, hence a path-connected open neighbourhood after restricting to a sufficiently small ball. Thus $X$ is locally path-connected, its path components are open, and connectedness forces there to be only one.

For the final example, putThe set $Y$ is path-connected: each vertical segment $C_n$ meets the horizontal segment $A$. The segment $B$ is also path-connected and lies in the closure of $Y$, because $(1/n,y)\to (0,y)$ for every $y\in [1/2,1]$. A connected set together with a connected subset of its closure is connected, so

It is not path-connected. The image of the first coordinate of any path in $X$ lies inbut, while the path has positive height, its first coordinate lies in the totally disconnected set $\{0\}\cup \{1/n:n\ge 1\}$. A path beginning on $B$ cannot leave $x=0$ before reaching height zero, and $X$ contains no point $(0,y)$ with $0\le y<1/2$. Hence no path joins $B$ to $Y$, and

Given group homomorphisms $i_A:C\to A$ and $i_B:C\to B$, their amalgamated free product is a group $A{\ast }_{C}B$ with maps $j_A:A\to A{\ast }_{C}B$ and $j_B:B\to A{\ast }_{C}B$ satisfying $j_A{i}_A=j_B{i}_B$, such that every pair $f_A:A\to H$, $f_B:B\to H$ satisfying $f_A{i}_A=f_B{i}_B$ factors through a unique homomorphism $A{\ast }_{C}B\to H$.

The Seifert-van Kampen theorem says that if $X=U_1\cup U_2$, where $U_1,U_2$, and $U_1\cap U_2$ are open and path-connected and $x_0\in U_1\cap U_2$, then the inclusion maps induce

Here is a direct proof of the requested generation statement. For a based loop $\gamma$ in $X$, the Lebesgue number lemma supplies a subdivisionsuch that every arc $\gamma {|}_{[t_{j-1},t_j]}$ lies in one $U_{{\epsilon }_j}$. Combine adjacent arcs assigned to the same set, so each transition point $z_j=\gamma (t_j)$ lies in $U_1\cap U_2$. Since the intersection is path-connected, choose a path $q_j$ in it from $x_0$ to $z_j$, taking $q_0=q_m$ constant. Thenis a loop in $U_{{\epsilon }_j}$. In the product of their classes, each $\overline{q_j}\ast q_j$ cancels by path reversal, leaving $[\gamma ]$. Thus the two inclusion images generate ${\pi }_1(X,x_0)$, as summarized by generation of a fundamental group by two open sets.

Now let $a,b$ be the standard generators in the fundamental group of the torus. A Möbius band retracts to a core circle, while its boundary traverses that core twice. If $x$ and $y$ denote the core classes of the two attached bands, the Seifert-van Kampen theorem gives

The two attaching classes $A=ab$ and $B=a^2{b}^3$ commute. Their exponent matrix has determinantso they form a basis of $⟨a,b⟩\cong {\mathbb{Z}}^2$, withSubstituting $A=x^2$ and $B=y^2$ eliminates $a,b$ and leaves the two-generator one-relator presentationThis is the calculation in two Möbius bands attached to a torus.

Finally send $x$ to $(12)$ and $y$ to $(23)$ in the symmetric group $S_3$. Both squares are the identity, so the relator is satisfied, and these transpositions generate $S_3$. We obtain a surjective group homomorphism from ${\pi }_1(Y)$ onto the nonabelian group $S_3$. Therefore ${\pi }_1(Y)$ is nonabelian.

No. TakeIt contains more than one point and has the fixed-point property of a closed interval, but it is not homeomorphic to the closed unit disc. Indeed, deleting any point $(t,0)$ with $0<t<1$ disconnects $B$, so such a point is a cut point. Deleting one point from a two-dimensional closed disc leaves a path-connected space. Since being a cut point is preserved by homeomorphisms, the two spaces cannot be homeomorphic. This gives the required counterexample.

Let $x,y\in \mathrm{Fix}(\varphi )$. A tree has a unique geodesic segment $[x,y]$. The isometry $\varphi$ sends this segment to the segment $[\varphi x,\varphi y]=[x,y]$. An isometry of a segment that fixes both endpoints fixes every point of it, soThus the fixed-point set of an elliptic isometry of a tree is a convex subtree, in particular it is path-connected.