$(\mathbb{Q}\times\mathbb{R})\cup (\mathbb{R}\times\{0\})$ is path-connected but not localy connected I have to prove that $A=(\mathbb{Q}\times\mathbb{R})\cup (\mathbb{R}\times\{0\})$ 
is path-connected, one in the chat suggested to take $$\varphi(t)=\begin{cases} (x,(1-3t)y), t\in [0,\frac13]\\ ((2-3t)x+(3t-1)x',0), t\in [\frac13,\frac23]\\ (x',(2t-2)y'), t\in [\frac23,1]\end{cases}$$ where $(x,y),(x',y')$ are two points from $A$. 
Is it sufficient to defined this continuous application to say that $A$ is path connected? can we for example defined $\varphi$ on $[0,\frac12]\cup [\frac12,1]$ ?
And how can i prove that $A$ is not localy connected, how to find a point from $A$ which has no local base of connected neighborhoods ?
Thank you.
 A: To "see" the path-connectedness, just notice that $A$ includes all the "vertical lines" ($Q \times R$) and one "horizontal line" ($R \times \{0\}$).
So any two points $(q_1,r_1)$ and $(q_2,r_2)$ are connected by the following continuous path: 


*

*first the "vertical segment" $(q_1,r_1) - (q_1,0)$

*second the "horizontal segment" $(q_1,0) - (q_2,0)$

*last the "vertical segment" $(q_2,0) - (q_2,r_2)$


That is basically what your function $\varphi$ does, but the geometric description may make it clearer to you.
The case when the second point (or both first and second) is of type $(r,0)$ is simple.
To show that $A$ is not locally-connected, you need to show that there exists a point $x$ with an open set $V$ st $x \in V \subset A$ and $V$ is not connected, i.e. you can divide $V$ into two disjoint open sets: $X,Y$ open, $V = X \cup Y$ and $X \cap Y = \{\}$. 
Advice:


*

*understand your topology (here the topology induced by $R^2$ on $A$), which will give you a description of your open sets (this is why the sets $(]q-e,q+e[ \times ]r-e,r+e[) \cap A$ are a basis of nbhds of $(q,r)$ in $A$) - don't overlook that in general topology exercises (!)

*visualize your set (here a "collection of disconnected vertical lines" plus a "horizontal line")

*check (and understand) your definitions of (i) connected, (ii) locally connected and (iii) path-connected

*remember that (path-connected) $\implies$ (connected) (but not the other way)

*then, consider a point $(q,r)$ "far enough" from the $(R \times \{0\})$ line) and how you can write $]0,2[_{\mathbb Q} = ]0,\sqrt2[_{\mathbb Q} \cup ]\sqrt2,2[_{\mathbb Q}$.

