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9

Neither is homeomorphic to $S^1$. $(-\infty,\infty]$ is not compact, while $S^1$ is, so they cannot be homeomorphic. $[-\infty,\infty]$ is compact, being homeomorphic to $[0,1]$, but removing any one point of $\Bbb R$ from $[-\infty,\infty]$ disconnects it, while removing one point from $S^1$ does not disconnect $S^1$, so they are not homeomorphic either.


7

There are many examples. The simplest is any space $X$ with more than one point that has the indiscrete topology, $\{\varnothing,X\}$. The simplest $T_1$ examples are any infinite set with the cofinite topology. The line with two origins is another $T_1$ example. Many of us topologists feel that there is no good reason to include Hausdorffness in the ...


7

I would argue that, overall in modern mathematics, the accepted meaning of "compact" is "every open cover has a finite subcover", i.e., it does not include the requirement of being Hausdorff. The approach of including Hausdorff-ness in the definition of compact, and instead using the word "quasi-compact" for the less restrictive condition, is traditionally ...


6

Yes. Any compact metric space is a continuous image of the Cantor set.


4

We'll prove the statement for $K$ compact metric space. Let $\epsilon >0$. The compact space $K$ can be covered by finitely many open balls $B(y_i, \epsilon/2)$, $i=1, \ldots m_{\epsilon}$. Therefore, there exist at most $m_{\epsilon}$ points in $K$, any two at distance at least $\epsilon$. Let $n_{\epsilon}$ be the largest number $n$ so that there exist ...


4

Let $f: \mathbb{N} \to [0, 1]$ such that the image of $f$ is dense, $\overline{f(\mathbb{N})} = [0, 1]$ (i.e. an enumeration of the rationals). There exists an extension of $f$ to the Stone–Čech compactification, $\overline{f}: \beta(\mathbb{N}) \to [0, 1]$. The image of $\overline{f}$ is compact and hence closed, yielding$$\overline{f}(\beta(\mathbb{N})) = ...


4

There is no right or wrong in these cases. I believe that Bourbaki has “quasi compact” for the non Hausdorff case, but the terminology can be seen elsewhere. The book where I learned topology is Kelley’s, where the Hausdorff property has its importance, but is not assumed throughout. Perhaps Engelking has his reasons for including the Hausdorff property ...


4

If $(-\infty,+\infty]$ is construed as a mere set rather than as a topological space, then it's not homeomorphic to anything including itself. Now lets add a topology: a basic open neighborhood of a point other than $+\infty$ is an open interval containing that point; a basic open neighborhood of $+\infty$ is a set of the form $(a,+\infty]\cup (-\infty,b)$; ...


3

For 2, take for example: $$X=\{a,b\}$$ $$\tau=\{\emptyset,X\}$$ Every finite space is compact, so $X$ is compact (according to your second definition), but $X$ is not Hausdorff because there is no open set that contains $a$ but not $b$. To include a less trivial example, let $(X,\tau)$ be any compact and Hausdorff space with more than one point. Now ...


3

Here's an approach using the universal property of the Stone-Čech-compactification: Let $(q_n)_n$ be an indexing of all rational numbers in the unit interval $I$, and let $f:\Bbb N\to I$ be a map sending $n$ to $q_n$. Then there is a unique map $f':\beta\Bbb N\to I$ such that $f'\iota = f$, where $\iota:\Bbb N\to\beta\Bbb N$ is the canonical embedding. Can ...


3

The Stone topology on $\beta X$ makes the principal ultrafilters $p_x$ isolated points for every $x \in X$, so if the canonical map $x \mapsto p_x$ is continuous (one of the properties of the Stone-Cech compactification), it forces $X$ to be discrete.


3

The following is an adaptation from Bourbaki's General topology, Ch. I, §10, Prop. 5. I will be using parts (a) and (c) of your exercise, namely: (a) if $f,g$ are universally closed and $\operatorname{Im} f$ is closed, then $g \circ f$ is universally closed; (c) if $g \circ f$ is universally closed and $g$ is injective then $f$ is universally closed. ...


2

Possibly the following proofs from “General Topology” by Ryszard Engelking would be helpful for you.


2

I agree that $F$ is bounded and countable. However $F$ is not closed. $\{0\}$ is a limit point that is not in $F$. $F \cup \{0\}$ is a compact set.


2

Let $x\in A$. Cover $A$ with the open cover $\bigcup_{n\geq 1}B_n(x)$ where $B_n(x)$ is a ball of radius $n$ centered at $x$. Then what does it mean to have a finite subcover?


2

I shall follow Exercise 3.12.5 from [Engelking1989]. A Hausdorff space $X$ is called H-minimal if every one-to-one continuous map of $X$ onto a Hausdorff space is a homeomorphism. I can show the following Proposition. Let $X$ be a Hausdorff space such that for any Hausdorff space $Y$ any map $f:Y\to X$ with closed graph is continuous. Then the space $X$ ...


2

He’s actually omitted a key observation: since $\sigma=\langle x_n:n\in\Bbb N\rangle$ has no convergent subsequence, it also has no constant subsequence, and therefore the set $S=\{x_n:n\in\Bbb N\}$ must be an infinite set. Thus, we might as well assume that the terms of the sequence are all distinct, i.e., that $x_m\ne x_n$ whenever $m\ne n$. Now we want ...


2

Others have already explained the differences in usage of various conventions. My first course in topology was taught by Engelking so perhaps I can say something about his motivations for this choice. He had very specific standards for notation and terminology. In particular, if I recall correctly, one guideline was that notions that are used more often ...


2

Related to some of the comments, you might be interested to know that $(-\infty,\infty]$ in the cyclic order topology is homeomorphic to $S^1$. The cyclic order topology is coarser than the corresponding linear order topology, in this case strictly coarser.


2

$A$ does indeed have a limit point $-$ in fact an infinite number of them. Every point on the unit circle is a limit point.


1

This is true in particular if $A = X$ is a topological manifold (or if $A$ is a submanifold of a manifold $X$, but the embedding has nothing to do here). I will sketch the proof and you feel free to relax the hypotheses as far as you can. I recall seeing this in my differential geometry course to prove the existence of a partition of unity. Lemma. Let $M$ ...


1

I guess a more general solution can be given as follows: Let $p$ be any polynomial, and $a$ be a constant real number. Then, since as $x$ goes to $\infty$ or $-\infty$ we know polynomial won't be bounded, we conclude that the set of $x$ satisfying $p(x)\leq a$ must be bounded. On the other hand, consider the set of $x$ such that $p(x)>a$. For each $x$ ...


1

This is a cool question and it's probably best to do it using sequences. First note that the quartic $q(x)=x(x^3-3x-1)-15=x^4+\ldots\,$ goes to $+\infty$ as $x\longrightarrow\pm\infty$, so the set $A$ of all $x\in\mathbb{R}$ with $q(x)\leq 0$ is bounded - this can be easily made rigorous. Instead, you could also use of the fact that ...


1

Markus Scheuer in his answer has analyzed the given proof and verified that it is correct. It cannot be denied that this proof has shortcomings, and above all, the introduction of the covering index function makes it more complicated than necessary. According to the standard definition a set $S\subset{\mathbb R}$ is compact if any given family ${\cal ...


1

Each of the sequence in the columns is finite/bounded, hence has a convergent subsequence by the Bolzano Weierstrass theorem. Then $(x_1^k)$ has a subsequence $(x_1^{n_k})$ that converges to, say, $x_1$, $(x_2^k)$ has a subsequence $(x_2^{n_k})$ that converges to $x_2$, and so on, $(x_n^k)$ has a subsequence $(x_n^{n_k})$ that converges to $x_n$. Let $x = ...


1

Hints:A metric space is sequentially compact if every sequence has a convergent subsequence. This is equivalent to every infinite subset of that space has a cluster point. A metric space being totally bounded is equivalent to the fact that every sequence has a Cauchy subsequence. Is it more obvious now?


1

No, you cannot deduce such a thing. For example, for $n=2$, you can take $K=[0,1]^2$. Then, take the point $x=(\frac12, 0)$. You can find that the line $t\mapsto (t, 0)$ for $t\in[0,1]$ (i.e. the line from $(0,0)$ and $(1,0)$) passes through $x$ and is contained entirely in $K$. Nonetheless, $x$ is not in the interior of $K$.


1

Let $A$ be a closed set in $X$. Let $a \in A$. Now, since $X$ is locally compact, we can find a compact neighborhood $V$ of $a$ in $X$. $V \cap A$ is closed in $V$, and closed subsets of compact sets are compact.


1

The space $\omega_1$ with the order topology is probably the simplest example. The simplest example that I know that does not require any knowledge of transfinite ordinals is the following one, which requires only a very basic knowledge of countability. Let $A$ be any uncountable set, and for each $\alpha\in A$ let $D_\alpha=\{0,1\}$ with the discrete ...


1

I think the "every convergent sequence with terms in the set $S=\{x1,x2,\cdots\}$ must be eventually constant" in the proof means all sequences with elements in $S$ that converge must be in this form: $$ x_{k_1}, x_{k_2}, \cdots , x_{k_n} , x_{k_n}, x_{k_n} , \cdots $$ Here you should treat $S$ as an ordinary set and we are talking about sequences whose ...



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