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A topological space $X$ is said to be locally connected at a point $x \in X$ if, for every neighborhood $U$ of $x$ (i.e. open set $U$ such that $x \in U$), there exists a connected neighborhood $V$ of $x$ such that $V \subset U$. If $X$ is locally connected at each of its points, then it is said to be locally connected.

Now is there any easy enough example of a connected space that fails to be locally connected at some point?

One example adduced by Munkres is the so-called topologist's sine curve, but I'm not sure why it is not locally connected.

Any other examples, please?

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4 Answers 4

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The intuition behind the several examples constructed for you here is as follows: take some space that you know to have many disconnected components that are 'very close' to each other. Examples include $\mathbb{Q}$ and $\{0,1/n, n\geq1\}$ (for which the disconnected components are the one-element subsets). These spaces are not locally connected because of the 'closeness' of the components: in the cases above, the space is not locally connected at $x = 0$ because no matter how far we zoom in on zero, we cannot isolate it from the connected components around it.

These spaces are of course not connected either, but we can make them so by joining all the components together somewhere far away. For instance, for the examples above, we could take the point $(0,1)$ in the plane (above the real line) and draw a line from this point to every point in our original space. This new space is connected because we can use these lines to move along a continuous path from any point to any other point. But it remains not locally connected since when we look at the vicinity of the points in the original space, we do not see this 'connection point' that we've added. All of the examples here operate along these lines.

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  • $\begingroup$ so nice of you! Your argument is much easier to grasp and is much more capable of being presented analytically as well! $\endgroup$ Commented Jan 3, 2018 at 10:54
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Another standard example that is even path connected but not locally connected is the comb space $$ [0,1]\times\{0\} ~\cup~ \{0\}\times[0,1] ~\cup~ \bigcup_{k\ge 1} \{\tfrac1k\}\times[0,1] $$

Every neighborhood of the point $(0,1)$ contains the top end of infinitely many of the teeth, but if the neighborhood is so small that it doesn't contain the base of the comb, it will be disconnected -- and making it smaller cannot possibly make it connected.

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  • $\begingroup$ thank you so much for your answer. Can you please supply some more details as to why this space is connected and why it is not locally connected? I'm sorry but I'm unable to figure out the details in your present answer. So I would appreciate if you could include some more detail or make your presentation more explicit. $\endgroup$ Commented Jan 2, 2018 at 20:56
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    $\begingroup$ It is connected because it's path connected. Given two points, you can form a path from one to the other by drawing a path from the point on your tooth to the base, then from the base to the base of the other tooth, and from the base of the tooth, back up to the point. I don't think I can clarify the reason the space is locally connected further; you should really start drawing this space, and draw some small circles around $(0, 1)$. $\endgroup$ Commented Jan 2, 2018 at 21:02
  • $\begingroup$ @HenningMakholm thank you for your reply. I think I can figure out why the comb space is connected, even without using the path-connectedness of this space. However, I'm still unable to conclusively figure out why no neighborhood of the point $0 \times 1$ that does not contain any point of $0 \times [0, 1]$ fails to be connected. $\endgroup$ Commented Jan 3, 2018 at 10:35
  • $\begingroup$ @SaaqibMahmuud: That was Duncan's reply. Suppose your neighborhood contains a ball of radius $r$ around $(0,1)$. Choose an integer $n>\frac1r$. Then the neighborhood contains a bit of the tooth at $\frac1n$. Now split your neighborhood into two sets depending on whether the $x$-coordinate is smaller or larger than $\frac12(\frac1n+\frac1{n+1})$. These sets are both open and nonempty. $\endgroup$ Commented Jan 3, 2018 at 15:28
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There is a standard example, let define the following set: $$\Gamma:=\overline{\{(x,\sin(1/x);x\in]0,1]\}},$$ then $\Gamma$ is connected, since it is the closure of a path-connected set but not locally connected, since in any neighborhood of zero, you see disjoint line segments.

Roughly, local-connectedness means that whenever you zoom in on a point, you still see a connected set.

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  • $\begingroup$ This is the "topologist's sine curve" mentioned in the question. $\endgroup$ Commented Jan 2, 2018 at 20:41
  • $\begingroup$ @HenningMakholm Indeed, but I tried to provide the missing insight on why it is not locally connected. $\endgroup$ Commented Jan 2, 2018 at 20:42
  • $\begingroup$ @C.Falcon how do we know that the line segments contained in any open ball centered at the point $0 \times 0$ are disjoint? You see, the curve $\left\{ \ x \times \sin \frac{1}{x} \ \colon \ 0 < x \leq 1 \ \right\}$ approaches asymptotically and continuously to the vertical line segment $0 \times [-1, 1]$. So how come the open ball will contain line segments that are disjoint? $\endgroup$ Commented Jan 3, 2018 at 10:40
  • $\begingroup$ You are right, but do you agree that in this process the line segment $\{0\}\times [-1,1]$ stays disjoint from $\{(x,\sin(1/x))\}$, there will always be a gap between the two. $\endgroup$ Commented Jan 3, 2018 at 12:42
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Let us consider the space $\mathbb R^2$.

The set $A=(\mathbb R\setminus\mathbb Q) \times\mathbb R$ is constituted of vertical parallel lines supported by irrational abscissas. This set is disconnected as you can have two open sets $]-\infty,q[\times\mathbb R\ ,\ ]q,+\infty[\times\mathbb R$ separating lines apart the rational abscissa $q$.

Let's add some connectivity in the horizontal direction by having $B=A\cup(\mathbb R\times\{0\})$.

Now $B$ is globally connected because there is a path between two points which starts with a vertical segment to reach the X-axis ($y=0$), an horizontal segment and finishes with another vertical segment.

Yet it is still locally disconnected because a neighbourhood of a point with a non-zero ordinate does not necessarily encounter the X-axis.

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  • $\begingroup$ thank you for your answer, but can you please check your post to see if some parentheses need to be inserted in order to make your presentation unambiguous? $\endgroup$ Commented Jan 3, 2018 at 10:42

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