topological properties of $X= \left( \cup_{n=1}^{\infty} \{\frac{1}{n}\} \times [0,1] \right) \cup (\{0\} \times [0,1]) \cup ( [0,1] \times \{0\})$

Question

Let $X= \left( \cup_{n=1}^{\infty} \{\frac{1}{n}\} \times [0,1] \right) \cup (\{0\} \times [0,1]) \cup ( [0,1] \times \{0\}) \subset \mathbb{R}^2$.

I want to show whether this space $X$ is normal and locally-connected, locally compact and even simply connected.

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Following are my trials :

First of all, I know $X$ is path-connected(hence connected), and compact.

(1) (path)-connectedness

Divdie $X=X_1 \cup X_2 \cup X_3$, then each of them are path-connected and their intersection is non-empty so their union is again path-connected].

(2) compact

Note in $\mathbb{R}^2$ by Heinel-Borel, compact=closed and bounded, clearly $X$ is bounded, and $X_1, X_2,X_3$ are closed so their union is closed

(3) Normal

I know compact Hausdorff space is normal. Since $X$ is compact and $R^2$ is Hausdorff, its subspace is again Hausdorff. so $X$ is compact Hausdorff and hence it is normal. [I can use the theorem "Closed subspace of normal is normal" and "metric space is normal"]

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About local properties.

Note that for Hausdorff space, I know compact implies locally compact, so $X$ is locally compact. But I am not sure about its locally connectedness, since local connectedness and connectedness are not related.

Furthermore, I am not sure about its simply connectedness. I know simply connected space is a path connected space with a trivial fundamental group, but I am not sure whether this space $X$ has trivial fundamental group.

Answer 1

The argument for compactness is incomplete, as we have an infinite union of closed sets so it's not clear it is closed).

But $S=\{0\} \cup \{\frac1n: n =1,2,3,\ldots\}$ is a convergent sequence so compact and $X$ can also be written as $S \times [0,1]$ (compact as a product of compacts) unioned with the compact $[0,1] \times \{0\}$, so compact in total.

It's metric (subspace of the plane) so normal as can be.

It's path-connected thanks to the subspace $[0,1]\times \{0\}$ (we can always find a path between points in different vertical stalks through there).

Local compactness is true (follows from compactness).

It's not locally connected at $(0,1)$ e.g. (any neighbourhood has clopen "stalk fragments" in it).

I think you can come with a contraction of $X$ to a single point (say $0,0)$) quite easily. So simple connectedness follows; it's homotopy equivalent to a point.

Answer 2

Henno Brandsma's answer is thorough enough, however I'd like to add something regarding simply connected: In order to show that $\pi_1(X) \cong 0$, we want to show that $X \simeq [0,1]$, i.e. $X$ is homotopy equivalent to $[0,1]$. We need to find maps $f : X \to [0,1]$ and $g: [0,1] \to X$ such that $$ f \circ g \simeq \text{id}_{[0,1]}\quad\text{and}\quad g \circ f \simeq \text{id}_X. $$ Let $g : [0,1] \hookrightarrow X$ be the inclusion map and let $f := \text{pr}_1$ be the projection of the first coordinate, i.e. $f(x_1,x_2) := x_1$. You can check that $(g \circ f)(x_1,x_2)=x_1$.

Next, define $H : X \times [0,1] \to X$ by $H((x_1,x_2),t) := (x_1,tx_2)$. We see that $H(-,0) = f(-)$ and that $H(-,1) = \text{id}_X(-)$, i.e. $H$ is a homotopy between $\text{id}_X$ and $g \circ f$. So we have that $X \simeq [0,1]$ and hence $\pi_1(X) \cong \pi_1([0,1]) \cong 0$, since $[0,1]$ is simply connected. Hope this was helpful to you!