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Topological properties


Active member
Jan 28, 2012
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.
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Well-known member
MHB Math Scholar
Feb 15, 2012
what you posted doesn't make a whole lot of sense to me.

first you state that $X = [0,1]$ is this a typo? i suspect you meant either:

a) $X = \Bbb R$ or:
b) $X = [-1,1]$ <--i think you mean this one.

it would be helpful to know which.

your proof that $\mathcal{T}$ is closed under finite intersections seems to be missing something:

what if $(-1,1) \subseteq A$ and $0 \not \in B$? and similarly for the unions? it seems to me you haven't considered "all the cases". i'm not saying it ISN'T a topology, i'm saying your proof is incomplete.

for (c) it is immediate that 0 lies in some set of an open cover. by the definition of your topology, this set must contain (-1,1). so you only need to add the open sets which cover {-1} and {1}, and you're done.

(d) looks ok, {-1,1} is clopen. thus for example:

$X = \{1\} \cup [-1,1)$

which is a disjoint union of two open sets.

i think it is first-countable: we can let the local base for $x \neq 0$ simply be $\{x\}$, as you suggest. for 0, we only have 4 neighborhoods:

(-1,1), [-1,1),(-1,1] and [-1,1], so we can take (-1,1) as a local base. the countability refers to the SET of neighborhoods, not the countability of the neighborhood set elements.

it is surely NOT second-countable, any base for $\mathcal{T}$ would surely have to contain $\{x\}$ for every $x \neq 0$ (we can't obtain these from unions), and there are uncountably many such singletons.

hmm...metrizable. it's clear that $X$ is $T_1$, but not Hausdorff. so no, not metrizable (we have separation failure for neighborhoods of 0).
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Active member
Jan 28, 2012
Thanks for the answers Deveno!
Indeed, I mean $X=[-1,1]$, that was a typo. Now, I'm aware of the fact that the neighbourhood base has to be countable not the element itself, thanks for pointing that.
I'll take a further look at the other cases to prove it's a topology.