# Check if $([0,1] \times \{0\})\cup \bigcup_{n=2}^\infty (\{\frac{1}{n}\} \times [0,1-\frac{1}{n}]) \cup ( [0,1] \times \{1\})$ is connected.

$$X = ([0,1] \times \{0\})\cup \bigcup_{n=2}^\infty \left(\left\{\frac{1}{n}\right\} \times \left[0,1-\frac{1}{n}\right]\right) \cup ( [0,1] \times \{1\})$$ Let $(X, d)$ be subspace of euclidean space. Check if $X$ is connected space.

$Y = X \setminus ( [0,1] \times \{1\})$ is connected (because it is sum of T-shape sets: $y_n = (\{\frac{1}{n}\} \times [0,1-\frac{1}{n}]) \cup ( [0,1] \times \{1\})$ and intersection $\cap_n y_n$ is not empty) and also $([0,1] \times \{1\})$ is connected.

There is also theorem that $S$ is connected if and only if for each nonempty sets $A$, $B$; $S=A \cup B$ $=>$ $\bar{A} \cap B \neq \emptyset$ or $A \cap \bar{B} \neq \emptyset$

Closure of $Y$ is $\bar{Y} = Y \cup (\{0\} \times [0,1])$ so $\bar{Y} \cup ( [0,1] \times \{1\}) = point(0,1) \neq \emptyset$ However is only one pair of $A$, $B$. How can I show it for each selection of A and B?

You could proof the following lemma (the proof is quite easy):

If $Y,Z \subset \mathbb R^n$ are connected subsets, such that there is a point $z \in Z$ with $d(Y,z)=0$, then $Y \cup Z$ is connected.

Use this with $Y = X \setminus ( [0,1] \times \{1\})$ and $Z=[0,1] \times \{1\}, z =(0,1)$ and you are done.

Note that in the lemma, we really need that point $z \in Z$, i.e. $d(Y,Z)=0$ is not enough as $(-1,0) \cup (0,1)$ shows.

I think the lemma is well known in the case, where the intersection is non-empty. This generalization shows that we do not need a point in the intersection, we just need that $Y$ gets arbitrary close to some point in $Z$.

• One more question. If we remove point $(0,1)$ form $Z$ then $Y \cup Z$ become not connected? – dyrAnd Feb 17 '16 at 10:42
• Yes. Then, around any point $z \in Z$, you can place some open epsilon-ball, which does not meet $Y$. Take $U$ to be the union of those balls, clearly an open set. Do the same for each point of $Y$ and you get an open set $V$, which covers $Y$ and does not meet $U$. – MooS Feb 17 '16 at 10:50