
#1
January 5th, 2013,
10:17
Hi, I need some help with the following question.
Let $ \displaystyle X=[0,1]$ 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.
Last edited by Siron; January 5th, 2013 at 10:33.

January 5th, 2013 10:17
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#2
January 5th, 2013,
11:29
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 firstcountable: 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 secondcountable, 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).
Last edited by Deveno; January 5th, 2013 at 15:01.

#3
January 5th, 2013,
11:51
Thread Author
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.