Locally connected spaces Let $X$ be a locally connected Tychonoff space without isolated points. If $f:X\rightarrow Y$ is a continuous bijection, then can we prove that $Y$ is locally connected (A space $X$ is locally connected if each $x\in X$ has a neighborhood base of open connected sets)? 
 A: The answer in general is no. For a simple counterexample, choose any bijection $\varphi: \mathbb{N} \to \mathbb{Q}$. Notice that $\varphi$ would already be a counterexample, except for the fact that $\mathbb{N}$ has isolated points. But we can easily fix that: we just take $X = \mathbb{N} \times \mathbb{R}$ and $Y = \mathbb{Q} \times \mathbb{R}$, and define $f: \mathbb{N}\times\mathbb{R}  \to \mathbb{Q}\times \mathbb{R}$ by $(n, x) \mapsto (\varphi(n), x)$. Now $f$ is a product of two continuous bijections and therefore a continuous bijection. Moreover, $\mathbb{N}\times \mathbb{R}$ has all properties we want, and $\mathbb{Q} \times \mathbb{R}$ is not locally connected. 

Since the previous example feels a bit like cheating the 'no isolated points' rule, let's give another counterexample, where this time we take $X$ to be an open half-open interval, let's say $[0, 1)$. For the space $Y$ we take the graph in $\mathbb{R}^2$ of $\sin(1/x)$ for $x \in (0, 1]$ and, we join it back to itself. This space is sometimes known as the Warsaw Circle. See the image below for a better desciption. The space $X$ is shown in blue and $Y$ is shown in red. The arrows are just to indicate the direction. Notice the importance of the fact that $X$ (the blue space) is open on the lower side, otherwise this wouldn't work. 

Now, clearly $X$ has all properties we want, but what about $Y$? Consider a point on the left of the red figure, somewhere at the height of the sine curve, and a small open ball around this point (see figure below). Then in this ball, we see the sine curve enter and leave infinitely often, so every small enough open neighbourhood of this point has infinitely many connected components. So this space is not locally connected at this point. 

