# Standard topology exercise with interior, compactness,connectedness

Let $$\tau^{*}$$ the subspace euclidian topology over $$[0,1]$$

We define $$\tau=\tau^{*} \cup$${$$\mathbb{R}$$}.

(i) Are $$\tau$$ finer or coarser of the standard euclidian topology $$\tau_e$$?

(ii) Find the interior of $$[0,+\infty)$$, $$[1/2,+\infty)$$ and $$\mathbb{Z}$$

(iii)Find $$\overline{[0,1]}$$

(iv)Is $$\mathbb{R},\tau$$ an Hausdorff space?

(v) Is it compact? Connected?

(i) I think the two topologies are not comparable since $$[0,1] \in \tau \setminus \tau_e$$ and $$(1,2) \in \tau_e \setminus \tau$$.

(ii) I found $$Int[0,+\infty)=[0,1]$$.

$$Int[1/2,+\infty)=(1/2,1]$$

And $$Int\mathbb{Z}=\emptyset$$

(iii) $$\overline{[0,1]}=\mathbb{R}$$

(iv) This space is not $$T_2$$ because every $$x \in [0,1]$$ is not Hausdorff separable from the points $$y \in (-\infty,0) \cup (1,+\infty)$$. In fact every open neighborhood of $$y$$ is the whole $$\mathbb{R}$$, so there aren't disjoint neighborhoods for $$x$$ and $$y$$, and the space is not $$T_2$$.

(v) It's compact because for every open cover, {$$\mathbb{R}$$} is a finite open cover for $$\mathbb{R}$$

It's connected because there aren't open, non empty, sets such that they are a partition of $$\mathbb{R}$$

Handy and obvious fact (that you already mentioned as well):

For any $x \in \mathbb{R}\setminus[0,1]$ the only open set containing $x$ is $\mathbb{R}$

This implies most of the following answers.

(i) is fine; (ii) are both correct, as no point outside $[0,1]$ be interior; (iii) is too, as all points outside $[0,1]$ are limit points of $[0,1]$; (iv) also implies that two points outside $[0,1]$ can never be separated.

As to (v): a little more argumentation would be nice:

$\mathbb{R}$ is compact, indeed, as for any open cover of $\mathbb{R}$, to cover $2$, e.g. must contain $\mathbb{R}$ as an element and then $\{\mathbb{R}\}$ is a very finite subcover.

connected: if $\mathbb{R} = U \cup V$ for open non-empty sets, then say $2 \in U$ and the fact implies that $U = \mathbb{R}$. So $U$ and $V$ are never disjoint. So $\mathbb{R}$ is connected in this topology.

• Thanks so much! ;)
– VoB
Aug 30, 2017 at 7:35