$$X = ([0,1] \times \{0\})\cup \bigcup_{n=1}^\infty (\{\frac{1}{n}\} \times [0,1]) \cup (\{0\} \times [0,1]) $$ 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?


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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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