# Local connectedness is not a continuous invariant

I am trying to prove the following: Show that local connectedness is not a continuous invariant by describing a continuous function $$f:[0,1] \rightarrow Y$$ from $$[0,1)$$ unto the space Y shown below.

Here is what I have done so far, but I am not sure whether this is rigorous enough:
Let the dot be point $$a$$ and let the curves coming from the stop at point $$b$$. $$[0,1)$$ is a locally connected space. Let $$f:[0,1] \rightarrow Y$$ be such that the path $$f$$ begins at $$a$$ and terminates at $$b$$. But then $$a$$ is not locally connected since any open set containing $$a$$ that also contains $$b$$ (that also does not contain the entire space of Y) does not contain any connected open sets. This is because that since $$a$$ and $$b$$ are so close that any open set containing $$a$$ must also contain $$b$$. Thus, local connectedness is not a continuous invariant.
Can anyone help me with making this proof more rigorous?
Thanks!

This question comes from Principles of Topology by Fred Croom, chapter 5.6, #12.

• Why doesn’t the text use $[0,1]$ in the discrete metric/topology? Any function is continuous with that domain and it’s locally connected trivially. Commented Apr 24, 2021 at 7:56

There is no point $$b$$: the wavy part of the curve does not have a left endpoint. You need to define $$f$$ piecewise, in five pieces. The first piece could be $$f(x)=\langle -5x,0\rangle$$ for $$0\le x\le\frac15$$: that maps $$\left[0,\frac15\right]$$ onto the segment of the $$x$$-axis from $$\langle 0,0\rangle$$ to $$\langle -1,0\rangle$$. The second piece could then map $$\left[\frac15,\frac25\right]$$ onto the vertical segment from $$\langle -1,0\rangle$$ down to $$\langle -1,-2\rangle$$:
$$f(x)=\left\langle-1,-10\left(x-\frac15\right)\right\rangle=\langle-1,2-10x\rangle\,.$$
The third piece could then map $$\left[\frac25,\frac35\right]$$ onto the horizontal segment from $$\langle-1,-2\rangle$$ to $$\langle 1,-2\rangle$$, the fourth then mapping $$\left[\frac35,\frac45\right]$$ onto the vertical segment from $$\langle 1,-2\rangle$$ up to $$\langle 1,\sin1\rangle$$, while the fifth could finish the job by mapping $$\left[\frac45,1\right)$$ onto topologist’s sine curve. For that last piece the map could be
\begin{align*} f(x)&=\left\langle 1-5\left(x-\frac45\right),\sin\frac1{1-5\left(x-\frac45\right)}\right\rangle\\ &=\left\langle5-5x,\sin\frac1{5-5x}\right\rangle\,. \end{align*}
I’ll leave it to you to work out the details for the third and fourth pieces and to convince yourself that the resulting function $$f$$ really does wrap the interval $$[0,1)$$ around the space $$Y$$, starting at $$\langle 0,0\rangle$$ (your $$a$$) and going counterclockwise around $$Y$$.
• @Chopin: You’re welcome. Since your approach involved the non-existent $b$, you’ll need to make some changes, but the idea of showing that $Y$ is not locally connected at $a=\langle 0,0\rangle$ is fine: if you take small enough nbhds, they have to contain disjoint nearly vertical pieces of the sine wave. (Note that I made a correction: the bottom horizontal line is at $y=-2$, not at $y=-1$.) Commented Apr 24, 2021 at 0:33