(a) Is a product of path connected spaces necessarily path connected?
(c) If $f:X \to Y$ is continuous and $X$ is path connected, is $f(X)$ necessarily path connected?
(d) If $\{A_\alpha \}$ is a collection of path connected subspace of $X$ and if $\bigcap A_\alpha \neq \emptyset$, is $\bigcup A_\alpha$ necessarily path connected?
My attempt: (a) let $\{ X_\alpha |\alpha \in A\}$. $X_\alpha$ is path connected, $\forall \alpha$. We need to show $(\prod_{\alpha \in A} X_\alpha ,\mathcal{T}_p)$ is path connected. Let $(x_\alpha), (y_\alpha) \in \prod_{\alpha \in A} X_\alpha$. Since $X_\alpha$ is path connected $\forall \alpha$, $\exists f_\alpha :[0,1]\to X_\alpha$ such that $f_\alpha$ is continuous and $f_\alpha (0)=x_\alpha$ and $f_\alpha (1)=y_\alpha$. Map $f:[0,1] \to \prod_{\alpha \in A} X_\alpha$ defined by $f(x)=(f_\alpha (x))_{\alpha \in A}$. $f_\alpha$ is continuous, $\forall \alpha$. By theorem 19.6, $f$ is continuous. $f(0)=(f_\alpha (0))_{\alpha \in A}=(x_\alpha)_{\alpha \in A}$ and $f(1)=(f_\alpha (1))_{\alpha \in A}=(y_\alpha)_{\alpha \in A}$. Thus $f$ is a path from $(x_\alpha)$ to $(y_\alpha)$. Hence $\prod_{\alpha \in A} X_\alpha$ is path connected. Is this proof correct?
(c) Example 4 & 5, Section 24 of Munkres’ Topology. Claim part.
(d) let $x,y \in \bigcup A_\alpha$. Then $x\in A_\beta$ and $y\in A_\gamma$. Since $\bigcap A_\alpha \neq \emptyset$, $\exists z\in A_\alpha$ $\forall \alpha$. So $z\in A_\beta ,A_\gamma$. Since $A_\beta ,A_\gamma$ is path connected, $\exists$ a path in $A_\beta$ from $x$ to $z$ and $\exists$ a path in $A_\gamma$ from $z$ to $y$. $\exists g:[0,1]\to A_\beta \subseteq \bigcup A_\alpha$ such that $g$ is continuous and $g(0)=x$ and $g(1)=z$. $\exists h:[1,2] \to A_\gamma$ such that $h$ is continuous and $h(1)=z$ And $h(2)=y$. $X=[0,1]\cup [1,2]=[0,2]$. By theorem 16.4, subspace topology, $\mathcal{T}_s$ on $[0,2]$ inherits from $\Bbb{R}$ is same as order topology, $\mathcal{T}_\lt$. $[0,1),(1,2] \in \mathcal{B}_{\lt} \subseteq \mathcal{T}_\lt$. So $[1,2]$ and $[0,1]$ are closed in $[0,2]$. $g(1)=h(1)=z$. Map $f:[0,2] \to \bigcup A_\alpha$ defined by $f(x)=g(x)$, $\forall x\in [0,1]$ and $f(x)=h(x)$, $\forall x\in [1,2]$. By theorem 18.3, $f$ is continuous. $f(0)=x$ and $f(2)=y$. Thus $f$ is a path from $x$ to $y$. Hence $\bigcup A_\alpha$ is path connected. Is this proof correct?