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In an Integral Domain is it true that $\gcd(ac,ab) = a\gcd(c,b)$?
How do I prove $\frac 34\geq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{n+n}$

The question is pretty much in the title, I’m looking for an example of a locally connected space and continuous mapping such that the image is not locally connected.

Thanks!

EDIT: Corrected the phrasing to the intended meaning.

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Consider the following variant on the topologist’s sine curve.

This space $X$ consists of the graph of $y = \sin(\pi/x)$ for $0<x<1$, together with a closed arc from the point $(1,0)$ to $(0,0)$. Note that $X$ is not locally connected at $(0,0)$.

However, there exists a continuous surjection $f\colon [0,2)\to X$. Specifically, $f(0) = (0,0)$ and $f(1) = (1,0)$, with $f(t)$ following along the bottom curve for $0\leq t\leq 1$. For $t>1$, the function follows along the sine curve, i.e.

$$

f(t) \;=\; \left(2-t,\sin\left(\frac{\pi}{2-t}\right)\right)\qquad\text{for }t> 1.

$$

Boring example: any space $X$ is the continuous image of the discrete topology on $X$ (using the identity and noting that any function with a discrete domain is continuous). A discrete space is trivially locally connected (all singleton sets). Now let $X$ be *any* non-locally connected space.

The graph-parametrisation of the graph of $\sin \frac{1}{x}$ for $x>0$ ? Instead of letting the graph trail off to the right as $x\to\infty$, just turn it around and let the curve run along the interval $[-1,1]$ on the $y$-axis. Then every point of this interval will have the property that local connectivity fails.

For example, consider x = NU{0} under discrete topology and Y = {0}U{1/n|n€N} as subspace of real line R.

Difine f:x——Y by f(0)=0,f(n)=1/n,n€N.then f is a bijection. and f is continues. Therefore,y is continous image of x. Note that x is being a discrete space is locally connected but y is not locally connected.

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