Is [0,1] under the subspace topology Hausdorff, Compact, or Connected?

[WWGD]

Ok, sorry, I am being lazy here. I am tutoring intro topology and doing some refreshers. Were given the subspace topology on [0,1] generated by intervals [a,b) and I need to answer whether under this topology, [0,1] is Hausdorff, Compact or Connected. I think my solutions work , but I am looking for alternatives in case student does not understand mine. Let X represent [0,1] on the Standard topology while X_new is [0,1] under the basis generated by [a,b) .
First, we see that intervals [c,d] ; $0\geq c< d\geq 1$ are open, by intersecting intervals $[a,x) \cap [0,1]$. It then follows that $[0,1]_New=[0,a) \cup [a,1]$ is disconnected. Hausdorff is relatively straightforward. Given$a,b \in [0,1] ; 0<a<b<1$ Then we can use $[0,a], [b,1]$( some accomodations when b=1)
It is compactness that seems harder. We have that X_new is strictly finer than X, since we saw (a,b) is open in X_new. Then the identity map from X_new to X is continuous, a continuous bijection.
Then all I have to show [0,1] is not compact in X_new is an argument by contradiction: A continuous bijection between compact and Hausdorff is a homeomorphism, which cannot happen, because the topologies X, X_ new are different . But this seems too high-powered an argument. Is there a simpler way of proving [0,1]_ new is not compact?

 

[Infrared]

It doesn't look to me that this topology is finer than the standard one. How do you show that $(1/2,1]$ is open, for example?

You also mention considering intervals $[c,d]$ where $c\leq 0$ and $d\geq 1$, which are not subspaces of $[0,1]$ if the inequalities are strict, so I think I'm misunderstanding something.

 

[WWGD]

I was thinking intervals $(a,c)$ are open. Let $a<b<c$. Then $b \in [\frac{b-a}{2}, \frac{c-b}{2})\subset (a,c)$, so that $(a,c)$ is open in this topology. This alone makes it weaker than the standard topology. So that the identity map between $[0,1]_new$ to $[0,1]$ is continuous, if I haven't botched anything. Then I may use , by contradiction, assuming$[0,1]_new$ is compact, that a continuous bijection between compact and Hausdorff is a homeomorphism. But this last seems too heavy-handed and I would prefer something simpler.

 

[Infrared]

I was thinking intervals $(a,c)$ are open. Let $a<b<c$. Then $b \in [\frac{b-a}{2}, \frac{c-b}{2})\subset (a,c)$, so that $(a,c)$ is open in this topology. This alone makes it weaker than the standard topology.

I don't think the open intervals $(a,b)$ generate the standard topology on $[0,1].$ You also need to include the intervals $[0,a)$ and $(b,1]$ to get the standard topology.

 

[WWGD]

I don't think the open intervals $(a,b)$ generate the standard topology on $[0,1].$ You also need to include the intervals $[0,a)$ and $(b,1]$ to get the standard topology.

Ok thanks, it's been a while since I've seen this material. [0,a) would follow by intersecting [0,1] with, e.g., (-1,a). Wouldn't we also get (b,1] by intersecting (b,2) with [0,1]?
But going back to compactness, do you have other ideas?

 

[Office_Shredder]

The space is not compact, consider
$$[0,1]= \bigcup_n [1/n,2)$$

 

[Infrared]

[0,a) would follow by intersecting [0,1] with, e.g., (-1,a). Wouldn't we also get (b,1] by intersecting (b,2) with [0,1]?

But $(b,2)$ isn't a subset of $[0,1].$

But going back to compactness, do you have other ideas?

I'm trying to understand what the topology is before giving an argument. But if $[a,b]$ (in particular singletons) are all open like you claim, then $\left\{ [1,1-1/n): n=2,3,...\right\}\cup\left\{ \{1\}\right\}$ is an open cover with no finite subcover.

@Office_Shredder The sets in your union are not subspaces of $[0,1].$

 

[Office_Shredder]

I assume from the posts above that the space is actually inheriting the topology induced from $\mathbb{R}$ that has the same opens sets.

And you're right, I wrote down the wrong thing. Your example is better.

 

[WWGD]

But $(b,2)$ isn't a subset of $[0,1].$I'm trying to understand what the topology is before giving an argument. But if $[a,b]$ (in particular singletons) are all open like you claim, then $\left\{ [1,1-1/n): n=2,3,...\right\}\cup\left\{ \{1\}\right\}$ is an open cover with no finite subcover.

@Office_Shredder The sets in your union are not subspaces of $[0,1].$

But , IIRC, for subspaces, a set is open if it is the intersection of a host space open set with the subspace. But you can also find basis elements to show either is open. For [a,c) ; 0<=a<c<1, just use a ball [0, (1-c)/2), so these are open. Similar for (a,1].
And I was hoping for an argument that used that [0,1]_new was weaker than [0,1], to show compactness.

 

[Infrared]

But , IIRC, for subspaces, a set is open if it is the intersection of a host space open set with the subspace.

Yes, this is the definition of the subspace topology.

But you can also find basis elements to show either is open. For [a,c) ; 0<=a<c<1, just use a ball [0, (1-c)/2), so these are open. Similar for (a,1].

I don't understand what you're writing here. Can you explain why $(a,1]$ is open in this topology, as it would have to be for this topology to be finer than the standard one?

I assume from the posts above that the space is actually inheriting the topology induced from $\mathbb{R}$ that has the same opens sets.

Right, I'm trying to figure out if this is the intended meaning, instead of literally the topology generated by the intervals $[a,b)$ in $[0,1].$

 

[WWGD]

Yes, this is the definition of the subspace topology. I don't understand what you're writing here. Can you explain why $(a,1]$ is open in this topology, as it would have to be for this topology to be finer than the standard one?Right, I'm trying to figure out if this is the intended meaning, instead of literally the topology generated by the intervals $[a,b)$ in $[0,1].$

Ok, my bad, I've been kind of sloppy. Let me be more clear from now on.

 

[Office_Shredder]

Ok, my bad, I've been kind of sloppy. Let me be more clear from now on.

But wait, which topology is it?

 

[WWGD]

But wait, which topology is it?

We start with the topology generated by the basis sets [a,b) on the Real line and then give [0,1] the subspace topology .

 

[Infrared]

We start with the topology generated by the basis sets [a,b) on the Real line and then give [0,1] the subspace topology .

Okay, that is not the topology you described in your OP! But anyway in this case I show non-compactness in post 7.

 

[Office_Shredder]

As an interesting point, the sequence $1/2-1/n$ does not converge to $1/2$ in this topology, since $[1/2,1)$ is an open neighborhood of $1/2$ which does not contain any elements of the sequence. Instead this sequence has no limit point.

 

[WWGD]

Just for the fun of it, let's check the metrizability of [0,1] under this topology. There are suficient conditions given is some metrization theorems IIRC. Nagata -Smirnov, Others. I only remember them vaguely.