0
$\begingroup$

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

  1. Connected

There is thorem that if $S$ is family of connected sets and $ \cap S\neq \emptyset $ then $\cup S$ is connected.

Line segments are connected in euclidean space and each of sets in $\bigcup_{n=1}^\infty (\{\frac{1}{n}\} \times [0,1]) \cup (\{0\} \times [0,1]) $ has one point common point with $([0,1] \times \{0\})$ so we can prove that finite union $$([0,1] \times \{0\})\cup \bigcup_{n=1}^{k} (\{\frac{1}{n}\} \times [0,1]) \cup (\{0\} \times [0,1])$$ is connected space.

But how can I show it for infinite union?

  1. Compact

I have no idea how to even start.

Any help please?

$\endgroup$
1
$\begingroup$

Show that each of the T-shaped (or L-shaped when $x=0$ or $x=1 $) sets $(\;[0,1]\times \{0\}\;)\cup (\;\{x\}\times [0,1]\;),$ (for $x\in N\cup \{0\}$), is connected. Their common intersection is $[0,1]\times \{0\},$ which is not empty, so their union $X$ is connected.

$\endgroup$
  • $\begingroup$ You can also apply the same theorem to show each of the T-shaped or L-shaped parts is connected : Each part is the union of a vertical and horizontal part,and each of these 2 parts is homeomorphic to [0,1], and they intersect. $\endgroup$ – DanielWainfleet Feb 18 '16 at 3:39
2
$\begingroup$

I think you might actually have an easier time showing that $X$ is path connected, and hence connected. Here's the strategy: Define a relation $x \equiv y$ if there is a path from $x$ to $y$, i.e. a continuous function $f : [0,1]$ with $f(0) = x$ and $f(1) = y$. This is an equivalence relation. Now all of the sets

$$\{0\} \times [0,1], \{1/n\} \times [0,1], [0,1] \times \{0\}$$

are path-connected. Try to show that for any $x,y \in X$, we have $x \equiv y$, hence $X$ is path-connected. Here's an example that should be illustrative in general:

Let $x = (1/3,1/2)$ and let $y = (2/3,0)$. Note that $$(1/3,0) \in (\{1/3\} \times [0,1]) \cap ([0,1] \times \{0\})$$ Now $\{1/3\} \times [0,1]$ and $[0,1] \times \{0\}$ are path-connected, so $x \equiv (1/3,0) \equiv y$. Hence $x \equiv y$.

For compactness, just note that any closed and bounded subset of Euclidean space is compact. Clearly $X$ is bounded. Showing it's closed is easy too, since it's a union of closed sets.

$\endgroup$
  • $\begingroup$ A union of closed sets may fail to be closed. In this case,it is easy to show that the complement of X is open. $\endgroup$ – DanielWainfleet Feb 17 '16 at 22:18
  • $\begingroup$ Right, I was confusing it with open, thanks! $\endgroup$ – Ethan Alwaise Feb 18 '16 at 1:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.