Is connectedness $\implies $ local connectedness? Let $X$ be a connected topological space.
Then I wonder that $X $ can be locally connected or not.
I think the title of my question is NOT correct considering a Euclidean space like $\mathbb R^n$.
Do you think so??
Also, conversely, if $X$ is a locally connected, and if some conditions are given additionally, Can $X$ be a connected space?
I think the latter case may be possible (not certainly), but don't know the necessary condition although it is true.
Please somebody tell me answers of them~~
 A: The Topologist's Curve is connected but not locally connected. Let $I_0, I_1\subseteq \mathbb{R}$ two disjoint intervals, $I_0\neq\emptyset\neq I_1$. Then the union $I_0\cup I_1$ is locally connected, but not connected.
A: Neither connected nor locally connected implies the other, nor do their negations.  Four examples:
$\Bbb R$ is connected and locally connected.
$[1,2] \cup [3,4]$ is locally connected but not connected
The topologist's sine curve is connected but not locally connected.
$\Bbb Q$ is neither
A: Many other answers have mentioned the topologist's sine curve as a counterexample. You may also be interested in the infinite broom, since the proof that it isn't locally connected is a bit easier. 
A: You can even take it to the next level: there exists a subspace $X \subset \Bbb{R}^2$ which is path-connected (so in particular connected), but at no point locally connected.
Namely, let 
\begin{split}
X&=\{(x,mx) \in \Bbb{R}^2 \; \vert \; 0 \leq x \leq 1, \; m \in [0,1] \cap \Bbb Q\} \\\\
& \bigcup \; \{(1-x,-mx) \in \Bbb{R}^2 \; \vert \; 0 \leq x \leq 1, \; m \in [0,1] \cap \Bbb Q\}
\end{split}
In concrete terms this means that $X$ consists of all line segments with rational slope between $0$ and $1$, connecting the point $(0,0)$ with the line $x=1$, and all line segments with rational slope between $-1$ and $0$, connecting $(1,0)$ with the line $x=0$  (try drawing an approximate picture and you will get what's going on).
You can now show that $X$ is indeed path-connected, but no point on it has a neighborhood basis consisting of connected sets. I leave it to you to flesh out the details of these two claims (but you can of course still ask for hints, if you don't succeed).
By the way (since credit should be to whom credit is due), I learned about this example right here.
