Вопросы из элементарной топологий

[vpb]

Bixel в сообщении #1641788 писал(а):

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 $[-1, 1]$?

This is correct.

To give you an example of accurate argument, I shall rewrite the latter argument in a very detailed way.

"Prove that $[-1,1]$ is a compact subset in topology $({\mathbb R}, \varepsilon_0)$. Suppose that $\{G_i\mid i\in I\}$ is a cover of $[-1,1]$ by open sets (here $I$ is some set, indexing elements of the cover). We have to prove that this cover contains a finite sub-cover covering $[-1,1]$.
As $0\in[-1,1]$, there exists an element of the cover, $G_i$, containing $0$. But the only open set in the topology $({\mathbb R}, \varepsilon_0)$, containing $0$, is ${\mathbb R}$. Therefore, $G_i={\mathbb R}$ for some $i\in I$. Since ${\mathbb R}\supset [-1,1]$, we see that the cover $\{G_i\mid i\in I\}$ contains a finite set of elements (in fact, a single element) covering $[-1,1]$. "

[Bixel]

vpb

Thank you for that nice write-up.

mihaild в сообщении #1641774 писал(а):

Space is locally compact if for any $x$ there is open $U$ and compact $K$ s.t. $x \in U \subseteq K$.

Is it always the case that $K \subseteq X$ in a topological space $(X, \tau)$?

I think $x \in \mathbb R \subseteq \mathbb R$ should work because $\mathbb R$ is compact in itself since an arbitrary open cover for $\mathbb R$ looks like $\{\mathbb R, G_2, G_3,...,\}$ and $\{\mathbb R\}$ definitely covers $\mathbb R.$

[mihaild]

Bixel в сообщении #1641800 писал(а):

Is it always the case that $K \subseteq X$ in a topological space $(X, \tau)$?

This is a strange question. We can ask about compactness only for subsets of topological space (and we need to specify which one).

Bixel в сообщении #1641800 писал(а):

an arbitrary open cover for $\mathbb R$ looks like $\{\mathbb R, G_2, G_3,...,\}$

Can you prove it?

[Bixel]

mihaild в сообщении #1641801 писал(а):

Can you prove it?

None of $G_i$ contain $0$ unless $G_i = \mathbb R$ and so every open cover must carry $\mathbb R$ for otherwise no open cover will cover $0$.

-- 08.06.2024, 02:56 --

Granted everything above is correct, let's move on to $(\mathbb Q, l_0).$ Let $S \subseteq \mathbb Q$. Then $T = \{\{x, 0\}: x \in S\}$ is an open cover. However $T$ has no finite subcover as it would miss some elements of $S$. Thus no point in $\mathbb Q$ has a compact neighborhood. Does it make sense?

[mihaild]

Bixel в сообщении #1641802 писал(а):

None of $G_i$ contain $0$ unless $G_i = \mathbb R$ and so every open cover must carry $\mathbb R$ for otherwise no open cover will cover $0$.

Yes. To give example of full proof:
Let $\{G_i | i \in I\}$ be open cover of $\mathbb R$. There is some $i$ s.t. $0 \in G_i$, then $G_i = \mathbb R$, and $\{G_i\}$ is finite subcover.

The last (simple) step is to show that for any point $x$ there is open set $U$ s.t. $x \in U \subseteq \mathbb R$.

Bixel в сообщении #1641802 писал(а):

However $T$ has no finite subcover as it would miss some elements of $S$

That's not necessary true.

[Bixel]

Long time no see. Sorry, I have so many competing demands. The whole thing fell by the wayside.

Anyway,

mihaild в сообщении #1641803 писал(а):

That's not necessary true.

Another possibility is to consider the set of all $[-x, x] \cup \{0\}.$ Then $\bigcup_{x = 1}^n [-x, x] \cup \{0\} = [-n, n] \cup \{0\}$, but $n + 1 \not \in [-n, n] \cup \{0\}$.

[mihaild]

$0 \in [-x, x]$ already.

Can you remind me, what do you want to prove? That $(\mathbb Q, L_0)$ isn't locally compact? For which point you want to prove that it doesn't have compact neighbourhood?

[Bixel]

mihaild в сообщении #1643231 писал(а):

That $(\mathbb Q, L_0)$ isn't locally compact?

Yes, thats exactly right. So for any $x \in \mathbb Q$ I believe $\cup [-x, x]$ covers $\mathbb Q$. But the finite union of these intervals would miss points like $x + 1$.

-- 19.06.2024, 01:53 --

If correct, this proof should also double as an argument for noncompactness of $(\mathbb Q, L_0)$.

[mihaild]

Bixel в сообщении #1643232 писал(а):

for any $x \in \mathbb Q$ I believe $\cup [-x, x]$ covers $\mathbb Q$

This is again meaningless: what variable do you take union by? If $x$ - then what is role of "for any $x \in \mathbb Q$"?

Also, to disprove local compactness, you need to show a point that has no compact neighbourhood. What point do you take?
(even if you think your proof works for any point, please choose some concrete, it will simplify reasoning)

[Bixel]

mihaild в сообщении #1643235 писал(а):

you need to show a point that has no compact neighbourhood. What point do you take?

If we show $\mathbb Q$ has no compact subset we are done.

Let $x$ be natural, $S \subseteq \mathbb Q$ and $T = S \cup \{0\}$. Then the union of all $(-x, x)$ covers $\mathbb Q$ and so $T$ is covered as well. Now $\bigcup_{x=1}^K (-x, x) = (-K, K)$ with $K+1 \not \in (-K, K)$. Thus $T$ has no finite subcover.

[mihaild]

Bixel в сообщении #1643237 писал(а):

If we show $\mathbb Q$ has no compact subset we are done.

This is impossible, because any finite subset is compact.

Bixel в сообщении #1643237 писал(а):

Thus $T$ has no finite subcover.

Why? It's not necessary that $K + 1 \in T$.
Consider, for example, $S = \{\pi, e\}$ in your construction.

[Bixel]

mihaild в сообщении #1643250 писал(а):

This is impossible, because any finite subset is compact.

This gives me an idea. Let $x$ be rational. Then we have $x \in \{0, x\} \subseteq \{0, x\}$. Since $\{0, x\}$ is a finite subset of $\mathbb Q$ and therefore compact it follows that $(\mathbb Q, L_0)$ is locally compact. The only problem is that $\{0, x\}$ is unordered.

[mihaild]

But we need not just compact set containing $x$, we need open set containing $x$ and contained in some compact set.

Bixel в сообщении #1643311 писал(а):

The only problem is that $\{0, x\}$ is unordered

Why is this a problem? What has order to do with this question?

[Bixel]

mihaild в сообщении #1643324 писал(а):

But we need not just compact set containing $x$, we need open set containing $x$ and contained in some compact set.

Isn't $\{x, 0\}$ both open and compact? Failing that, $x \in \mathbb Q \subseteq \mathbb Q$ should do the trick provided $\mathbb Q$ is compact in the given space.

mihaild в сообщении #1643324 писал(а):

Why is this a problem? What has order to do with this question?

I figured since $\mathbb Q$ is ordered we want all its subsets ordered as well. Guess not.

[Bixel]

I suspect $\mathbb Q$ is not compact. Will prove it later. No infinite subset of rationals is compact. Will have to prove this, too. Then if $K$ is a compact subset of rationals, then $\frac 15 \in [0, \frac 15] \not \subset K.$