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Hi, I need some help with the following question.

Let [tex]X=[0,1][/tex] en let $\mathcal{T}$ be a topology on $X$ defined as

$$\mathcal{T}=\{U \subset X | ]-1,1[\subseteq U \ \mbox{or} \ 0 \notin U\}$$

Answer the following questions:

(a) Does $\mathcal{T}$ defines a topology on $X$?

It's clear that $X \in \mathcal{T}$ because $]-1,1[ \subset X$. Also $\emptyset \in \mathcal{T}$ because $0 \notin \emptyset$. Suppose $A,B \in \mathcal{T}$, $0 \notin A$ and $0 \notin B$ then $0 \notin A \cap B$ therefore $A \cap B \in \mathcal{T}$. This last argument can also be used to prove that $\mathcal{T}$ is closed under infinite unions.

(b) Is $X$ a compact topological space?

I think it's compact. Let $(O_j)_j$ be an open cover of $X$. Now, take one open set $O_k$. If $0 \in O_k$ then $]-1,1[ \subset O_k$, then $O_k$ almost covers $X$ except (in the sadest case) $-1$ and $1$, but because $(O_j)_j$ is a cover of $X$ there must be open sets $O_1$ and $O_{-1}$ who cover $1$ and $-1$ thefore $O_k \cup O_1 \cup O_{-1}$ is a finite subcover.

If $\forall O_i \in (O_j)_j: 0 \in O_i$ then there's a finite subcover.

It's impossible that $\forall O_i \in (O_j)_j: 0 \notin O_i$ because then it wouldn't be a cover thus there must always be an open set which includes $0$.

This is why I conclude $X$ is compact. I know it's a mess but maybe someone could give me a better and more structural proof if I'm correct.

(c) Is $X$ connected?

I think $X$ is not connected, because in my opinion $\{-1,1\} \subset X$ is open because $0 \notin \{-1, 1\}$, but it's also closed because it's the complement of $]-1,1[$ which is open.

(d ) Is $X$ first countable and second countable?

I can't prove this statement. I have some thought's but I can't structure them into one proof. First, I was thinking, maybe I can take as a countable neighbourhoud for every $x \in X$ just $\{x\}$, but that's not possible for $0$. Because every neighbourhoud of $0$ thus also every neighbourhoud base has to include $]-1,1[$ which is not countable. Hence it's not first countable. But I'm not sure of this.

(e) Is $X$ metrizable?

I have no idea how to prove this. I thought maybe I could try to prove the topology is Hausdorff, but I don't think it would follow from there.

Let [tex]X=[0,1][/tex] en let $\mathcal{T}$ be a topology on $X$ defined as

$$\mathcal{T}=\{U \subset X | ]-1,1[\subseteq U \ \mbox{or} \ 0 \notin U\}$$

Answer the following questions:

(a) Does $\mathcal{T}$ defines a topology on $X$?

It's clear that $X \in \mathcal{T}$ because $]-1,1[ \subset X$. Also $\emptyset \in \mathcal{T}$ because $0 \notin \emptyset$. Suppose $A,B \in \mathcal{T}$, $0 \notin A$ and $0 \notin B$ then $0 \notin A \cap B$ therefore $A \cap B \in \mathcal{T}$. This last argument can also be used to prove that $\mathcal{T}$ is closed under infinite unions.

(b) Is $X$ a compact topological space?

I think it's compact. Let $(O_j)_j$ be an open cover of $X$. Now, take one open set $O_k$. If $0 \in O_k$ then $]-1,1[ \subset O_k$, then $O_k$ almost covers $X$ except (in the sadest case) $-1$ and $1$, but because $(O_j)_j$ is a cover of $X$ there must be open sets $O_1$ and $O_{-1}$ who cover $1$ and $-1$ thefore $O_k \cup O_1 \cup O_{-1}$ is a finite subcover.

If $\forall O_i \in (O_j)_j: 0 \in O_i$ then there's a finite subcover.

It's impossible that $\forall O_i \in (O_j)_j: 0 \notin O_i$ because then it wouldn't be a cover thus there must always be an open set which includes $0$.

This is why I conclude $X$ is compact. I know it's a mess but maybe someone could give me a better and more structural proof if I'm correct.

(c) Is $X$ connected?

I think $X$ is not connected, because in my opinion $\{-1,1\} \subset X$ is open because $0 \notin \{-1, 1\}$, but it's also closed because it's the complement of $]-1,1[$ which is open.

(d ) Is $X$ first countable and second countable?

I can't prove this statement. I have some thought's but I can't structure them into one proof. First, I was thinking, maybe I can take as a countable neighbourhoud for every $x \in X$ just $\{x\}$, but that's not possible for $0$. Because every neighbourhoud of $0$ thus also every neighbourhoud base has to include $]-1,1[$ which is not countable. Hence it's not first countable. But I'm not sure of this.

(e) Is $X$ metrizable?

I have no idea how to prove this. I thought maybe I could try to prove the topology is Hausdorff, but I don't think it would follow from there.

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