Вопросы из элементарной топологий

[mihaild]

Bixel в сообщении #1641704 писал(а):

because $\mathbb R$ is an open neighborhood of $0$?

Because $\mathbb R$ is the only one open neighborhood of $0$. So, any open neighbourhood of $0$ contains all terms of $x_n$, and thus $x_n \to 0$.

Bixel в сообщении #1641704 писал(а):

Since $U \in l_0$ and $\{x_n\}$ is rational $\{x_n\} \subseteq U$

Why?

[Bixel]

mihaild в сообщении #1641705 писал(а):

Why?

I think I see what you're alluding to. Consider $\{0\}$. It is open in $l_0$, but contains no $x_n$for any $n$. For $\{x_n\}$ to converge to $0, \ x_n$ must be in every open set for all but finite number of terms. That means $(\mathbb Q, l_0)$ is not sequentially compact. Make sense?

[mihaild]

Bixel в сообщении #1641706 писал(а):

That means $(\mathbb Q, l_0)$ is not sequentially compact. Make sense?

No, that just means there is sequence not converging to $0$. To prove $(\mathbb Q, l_0)$ is not sequentially compact, you need to find a sequence s.t. no it subsequence converges (not just to $0$, to anything).

[Bixel]

mihaild в сообщении #1641707 писал(а):

To prove $(\mathbb Q, l_0)$ is not sequentially compact, you need to find a sequence s.t. no it subsequence converges (not just to $0$, to anything).

What about $\{n\}$ where $n$ is natural? It's a subset of rationals and has no convergent subsequence.

[vpb]

Bixel в сообщении #1641686 писал(а):

But I guess since the open sets are not ordered we could just the drop the indexing.

Yes, correct.

Bixel в сообщении #1641686 писал(а):

By $\{x_n\} \setminus \{0\} \in \mathbb Q$ I meant to denote a rational sequence missing $0$. Maybe $\{x_n\} \cap \mathbb Q \setminus \{0\}$ would've worked better.

No, this formula makes no sense also. As the colleague said already, the correct formula is $x_n\in{\mathbb Q}\setminus\{0\}$, or $\{x_n\}\subseteq {\mathbb Q}\setminus\{0\}$,
or $\{x_n\mid n=1,2,\ldots\}\subseteq {\mathbb Q}\setminus\{0\}$. But to avoid any possible misunderstandings, usually a mix of symbolic and natural language is used, for example " Let $(x_n)=(x_1,x_2,\ldots)$ be a sequence of rational numbers different from zero", or "$(x_n)$ be a sequence of rational numbers
such that $x_n\ne 0$ for all $n$", and the same can be expressed in many other ways also.

Of course, the rest of the proof also should be written clearly.

I would like to inform you about several more elementary books on topology, they may be useful for you. These books are written in a more classical style, with more details, than McCluskey.
W.G.Chinn, N.E.Steenrod, First concepts of topology
B.Mendelson, Introduction to topology
S.A.Morris, Topology without tears

(These books may be found at the so called "libgen" cite, which is now at url "libgen.rs". But this is a pirate resourse, so it may change its location).

-- 07.06.2024, 03:08 --

Bixel в сообщении #1641712 писал(а):

and has no convergent subsequence.

This is correct, but this should be proved.

[Bixel]

vpb

Duly noted :) Appreciate the suggestions. Thanks.

[Bixel]

Let $x \in \mathbb Q, y \in \mathbb R.$ Then $[x - 1, x +1], [y - 1, y + 1]$ are a compact neighborhood around any rational and real, respectively, meaning $(\mathbb Q, l_0), (\mathbb R, \varepsilon_0)$ are both locally compact. Correct?

[mihaild]

Space is locally compact if for any $x$ there is open $U$ and compact $K$ s.t. $x \in U \subseteq K$. Which of $U$ and $K$ do you call "compact neighborhood"?
Also, the important part is not to guess answer (there are only two possibilities, so with two attempts it's easy), but to prove it. Choose one question and try to answer it with a proof.

[Bixel]

Lets focus on $(\mathbb R, \varepsilon_0)$. Reals are open in this space which means for any $x \in \mathbb R$ we have $x \in \mathbb R \subseteq [-\infty, \infty]$. Not sure if abusing the extended real line like this is legal.

[mihaild]

$K$ above should be a compact subset (not necessary proper) of $\mathbb R$. Without any "extensions" (extension can change local compactness).

Lets start with something simple. Can you find few examples of compact sets in $(\mathbb R, \varepsilon_0)$?

[Bixel]

$[-1, 1]$ for one. We can come up with a few more in a similar manner.

Let $x \ne 0$ be real. Then $(x - 1, x + 1) \subset [x - 1, x + 1]$ should work.

If $x = 0$, though, then $\mathbb R$ is the only open set that contains $x$.

[mihaild]

Bixel в сообщении #1641780 писал(а):

$[-1, 1]$ for one

Can you prove that $[-1, 1]$ is compact in $(\mathbb R, \varepsilon_0)$?

Bixel в сообщении #1641780 писал(а):

Then $(x - 1, x + 1) \subset [x - 1, x + 1]$

$(x - 1, x + 1)$ is not necessary open.

[Bixel]

mihaild в сообщении #1641781 писал(а):

Can you prove that $[-1, 1]$ is compact in $(\mathbb R, \varepsilon_0)$?

$0 \in [-1, 1]$ and so it's closed. It's also bounded. Though this theorem might work in the metric spaces only. It's getting to be too complicated :) Scratch that. Lets try something else. Let $[-1, 1] \subseteq \cup G \in \varepsilon_0.$ Let $G_1 = (-5, 0), G_2 = (0, 5), G_3 = \mathbb R$ and together they should cover $[-1, 1].$

[mihaild]

Bixel в сообщении #1641784 писал(а):

It's also bounded. Though this theorem might only work in the metric spaces

In generic topological space there is no notion of "boundness", it requires either metric or vector structure.

Bixel в сообщении #1641784 писал(а):

Let $[-1, 1] \subseteq \cup G.$ Let $G_1 = (-5, 0), G_2 = (0, 5), G_3 = \mathbb R$ and together they should cover $[-1, 1].$

No, set being compact means that any open cover has finite subcover.

[Bixel]

mihaild в сообщении #1641785 писал(а):

No, set being compact means that any open cover has finite subcover.

Since $\mathbb R$ is the only open set that can cover $0$, any open cover of $[-1, 1]$ would necessarily have to contain $\mathbb R$. But then $\mathbb R$ alone is enough to cover $[-1, 1]$?