# Are the following subsets open in the standard topology?

1. Feb 4, 2017

### sa1988

1. The problem statement, all variables and given/known data

Determine whether the following subsets are open in the standard topology:

a) $(0,1)$
b) $[0,1)$
c) $(0,\infty)$
d) $\{x \in (0,1) : \forall n \in \mathbb{Z}^{+}$ $, x \not= \frac{1}{n}\}$

2. Relevant equations

3. The attempt at a solution

a) $(0,1)$ is open because for any $x$ in the given interval $0 < x < 1$, it always has reachable neighbourhoods without any boundary.

b) $[0,1)$ is not open because the point $x=0$ has neighbourhoods outside the set. The set is also not closed because its complement gives $(-\infty , 0]$ and $(1,\infty )$ where the former is not open for similar reasoning.

c) $(0,\infty)$ is open, because all values $x$ in $0 < x < \infty$ have neighbourhoods, similar to part a).

d) $\{x \in (0,1) : \forall n \in \mathbb{Z}^{+}$ $, x \not= \frac{1}{n}\}$ is not open because there are discontinuities every time one reaches a point $\frac{1}{n}$. For this reason the set is also not closed, because the complement will leave 'opposite' discontinuities.

As with my previous thread on basic set questions, I'm hoping I've got these right but am running it by the forum because Topology and set theory in general is still currently fairly new territory for me.

Many thanks.

2. Feb 4, 2017

### Orodruin

Staff Emeritus
You may want to reconsider (d).

Ask yourself if for any given $x$ in the specified set there exists a neighbourhood that is within the set.

3. Feb 4, 2017

### sa1988

Hmm, I think I know what you mean.

The set has all $\frac{1}{n}$ removed, but it's possible to get infinitely close to that 1/n without ever reaching it, i.e. the point at $\frac{1}{n} - \frac{1}{9999999...}$

So the discontinuities are there, but the 'limit' is unreachable. All $x$ in the set have neighbourhoods. Thus the set is open.

I would then say that the set is still not closed, because its complement gives only the specific values $\frac{1}{n}$, none of which have neighbourhoods.

4. Feb 4, 2017

### Orodruin

Staff Emeritus
Right. Alternatively, the set can be written as
$$\bigcup_{n=1}^\infty \left(\frac{1}{n+1},\frac 1n\right),$$
which is a union of open sets.

5. Feb 4, 2017

### PeroK

For a set not to be open there must be some point that does not have a neighbourhood within the set. In question b) you identified such a point. To show this set is not open, you must find such a point. A narrative about "discontinuities" does not suffice.

6. Feb 4, 2017

### sa1988

In my hand-written homework response I've written that the set is open, "... because all $x$ exist in an open interval $\frac{1}{n+1} < x < \frac{1}{n}$ "

To illustrate the set not being closed, I wrote that the complement gives all the points $\frac{1}{n}$, all of which have no neighbourhoods.

I'm not sure if that latter sentence is good enough, however. I'm aware that my weakness is in turning 'wordy' answers into mathematical notation.

7. Feb 4, 2017

### PeroK

Post #4 gives one option, which is probably the best if you are allowed to assume this property of open sets. Alternatively, you have to find a neighbourhood for every $x$, based on the intervals you have identified.

8. Feb 4, 2017

### Orodruin

Staff Emeritus
Any union of open sets is an open set by definition of a topology.

9. Feb 4, 2017

### PeroK

Yes, very true.

10. Feb 4, 2017

### sa1988

Thanks for the responses - advice noted.

11. Feb 5, 2017

### micromass

Staff Emeritus
And not by definition for open sets in $\mathbb{R}^n$, in which the OP is working. In that case, it actually needs a proof. And if it's not proven, the OP cannot use it.

12. Feb 5, 2017

### sa1988

Forgive me if I'm mistaken but does my OP not state that we're working with the standard topology, rather than $\mathbb{R}^n$?

If not, then I have another hurdle in this wonderful quagmire I have waded into known as set theory and topology. Ha.

13. Feb 5, 2017

### micromass

Staff Emeritus
Sure, but you actually need to prove that the standard topology on $\mathbb{R}$ satisfies the axioms of a topology! Once you did that, it's fine. But you can't just use the axioms without verifying them, like Orodruin wanted to do.

14. Feb 5, 2017

### sa1988

My biggest downfall in the whole area of mathematics is proofs. For my physics degree it's been mostly about "modelling the world with maths", but now I'm in my final year and half my courses are taught directly by the maths department. Suddenly abstract, rigorous proofs are the done thing.

I shall soldier on!

15. Feb 5, 2017

### PeroK

In addition to rigorous proofs, the other difference with pure maths is that you can only assume what you have already proved. In applied maths, on any problem you can (more or less) cherry pick from all the maths you've ever learned. With pure maths you cannot, for example, assume that complex numbers exist and use their properties if you are doing real analysis.

In this case, it may be relevant what your course defines as the standard topology, where your starting point was and what you can assume you have already proved when tacking these problems.

16. Feb 5, 2017

### sa1988

This is another thing I'm needing to hurdle - the parts that say, "This is true because of proposition 1.34 from earlier in the notes", where the proof for that proposition is a thing I need to read about 50 times because it's in a style or language which is still quite new to me.

I guess this is where mathematicians can push out the smug banter towards physicists who don't do 'proper maths'.

17. Feb 5, 2017

### Orodruin

Staff Emeritus
Well, the standard topology is called a topology for a reason and referred to as such in the OP so I implicitly inferred from this that this has already been shown in whatever text or course the OP is following and can be assumed.

In fact, it would make no sense to refer to open sets before you have shown that they form the open sets of a topology.

That you need to show that the standard topology is actually a topology before you call it a topology should go without saying. Obviously I do understand that one has to show that a topology is a topology before using its properties and I am definitely not taking it for granted.