# Tag Info

9

But how can we infer the existence of an unbounded continuous on the space knowing only about some sequence of points in this space By using the sequence to construct an unbounded continuous function. Since the sequence - call it $(x_n)_{n\in\mathbb{N}}$ - has no convergent subsequence, every point occurs only finitely many times in the sequence. ...

6

The claim is true if $Y$ is Hausdorff: if $C\subseteq Y$ is compact, then it is closed; therefore $f^{-1}(C)$ is closed in $X,$ hence compact. For a counterexample, take $Y=X$ with the indiscrete topology and $f$ the identity map. Then every subset of $Y$ is compact, which can be easily arranged for $X$ not to.

6

I think you are reading the wrong part of your book. That definition (top of p. 185 in my edition) is immediately preceded by a Theorem 29.1, which is a long explanation of how one can take a noncompact locally-compact Hausdorff space $X$ and embed it into a compact space $\def\Y{X^\ast}\Y$ that has exactly one point more than $X$. This larger space $\Y$ ...

5

If a function $f:X\to Y$, between any topological spaces, has finite image, then it maps any set in $X$ to a compact set in $Y$, and in particular thus it preserves compactness. Further, if $X$ is totally disconnected, so every connected component is a singleton, then the image of $f$ on any connected set in $X$ is a singleton too, and thus connected in $Y$. ...

5

As you said, it is bounded since $C\subseteq [0,1]^n$. To see it's closed notice that $$F\colon\ \mathbb R^n\to\mathbb R,\ (x_1,\dots,x_n)\mapsto \sum_{i=1}^n x_i$$ is a continuous map. Since $\{1\}\subseteq \mathbb R$ is closed it follows that $F^{-1}(1)\subseteq \mathbb R^n$ is closed as well. Now you have $$C = F^{-1}(1) \cap [0,1]^n$$ which is ...

4

You can say the following: consider a sequence $(x_n)$, which has no convergent subsequence. Then, inductively, you can construct a sequence of positive numbers $\varepsilon_n$ such that $\overline{B(x_n,\varepsilon_n)}\cap\overline{B(x_m,\varepsilon_m)}=\emptyset$ for all $n\neq m$. Then, define a function $f$ as follows: $$f(x)=\left\{\begin{array}{c ... 4 The curve is the image of a continuous function over the set [0,1] as you can see in the answer that you post. Therefore, as compactness is preserved under continuous functions, the Koch curve is compact. 4 Take X=(0,\infty) with the usual metric. [1,2] is a closed, bounded and compact set in X. (0,1] is a closed and bounded set in X, which is not compact (e.g. (0,1]\subseteq\bigcup_n(1/n,2)). [1,\infty) is a closed, but unbounded and not compact set in X. (1,\infty) is an unbounded set which is neither closed nor compact in X. (1,2) is ... 4 It is not necessary for H to be normal or have any other property except that it is a subgroup, but then G/H, in general, is not a group any more. However, it is always a set, and it becomes a topological space when equipped with the quotient topology with respect to the canonical map \pi\colon G \to G/H, which is then continuous, surjective, and open. ... 4 Yes, \{x\} is a set which has a single element (and thus we call it a singleton), and \{x\}\times B=\{(x,b)\mid b\in B\}. Clearly it is homeomorphic to B. Note that ordered pairs are not necessarily ordered pairs of real numbers. You can talk about product of two sets, or two spaces. And this is somewhat similar to the case of \Bbb R^2, or the real ... 4 An equivalent definition of compactness is the following: A space X is compact if and only if every family of closed subsets of X with the finite intersection property has non-empty intersection. We say that a family \mathcal F of sets has the finite intersection property if F_1\cap\cdots\cap F_n\ne\varnothing, for every n and ... 3 This is not true in general. Let X=Y=[0,1]. Take X with the usual topology. For Y, take the topology$$\tau=\left\{\varnothing,Y,(1/2,1]\right\}.$$Then id:x\in X\mapsto x\in Y is continuous, but (1/2,1]=id^{-1}(1/2,1] is not compact, although (1/2,1] is compact in Y. On the other hand, if Y is Hausdorff, then every compact of Y is ... 3 Proposition Let X be a topological space with the Bolzano Weiertrass property. Then every countable covering of X admits a finite subcovering. Proof Let \{O_1,O_2,\ldots\} be the countable open cover of X, so that X\subseteq\bigcup O_n. Suppose to the contrary that no finite collection of the cover covers X. Then in particular ... 2 Very broad hint: You need to prove three things with the hint to apply the theorem and show that A \neq \emptyset. Each of them is provable by induction, I'll let you write down the details. \color{red}{\forall n, A_{n+1} \subset A_n}: f(X) \subset X \Rightarrow f(f(X)) \subset f(X) \Rightarrow f(f(f(X))) \subset f(f(X))... \color{red}{\forall n, ... 2 The definition of compactness means that for any open cover \mathcal{U} of the space X there are finitely many U_1 , \ldots , U_n from that collection \mathcal{U} which also covers X. When you "add X" to \mathcal{U}, you are changing the open cover into a different open cover, let's call it \mathcal{U}^\prime. While the collection \{ X , ... 2 The basic idea is correct (taking complements and using de Morgan, essentially). As suggestions for write-up: show the directions, for left to right e.g.: Suppose X is compact. Let \{ C_j: j \in J \} be a collection of closed sets with empty intersection. Then define, for each j \in J, U_j = X \setminus C_j, which is open in X. Then$$\cup_{j ...

2

By definition of sequential compactness, we must show that every $(y_n) \subseteq f(S)$ has a convergent subsequence. To this end, pick $(y_n)$ in $f(S)$ arbitrary. Since $y_n \in f(S)$, then $y_n = f(x_n)$ for some $x_n \in S$. By sequential compactness of $S$, $(x_n)$ contains a subsequence $x_{n_k}$ that converges in $S$, say $x_{n_k} \to l \in S$. ...

2

Let $X$ be a topological space. A closed set $A\subseteq X$ is a set containing all its limit points, this might be formulated as $X\setminus A$ being open, or as $\partial A\subseteq A$, so everypoint in the boundary of $A$ is actually a point of $A$. This doesn't mean $A$ is bounded or even compact, for example $A=X$ is always closed. If $X$ is a metric ...

2

Suppose you do something funny, and consider a base $\mathcal{B}$ of a topological space $X$ with $X \in \mathcal{B}$. Clearly there are covers of $X$ by sets in $\mathcal{B}$ admitting a finite subcover: all of them which include $X$, for example. But not all topological spaces are compact, so something must be amiss. To see this, you really should be a ...

1

Finding a cover with no finite subcover is the right way to go. But I think that you are distracting yourself by trying to make the elements of your cover pairwise disjoint: this is not necessary. (It's not wrong either - you're just adding a complication.) Anyway, as @Omnomnomnom notes, $[1,2)$ is open in $\langle\mathbb{R},\tau\rangle$, so if you can ...

1

My suggestion: Let $T$ be the topological space. Let $U_{i}\neq\emptyset, i\in I$ be an open cover. Let $i_{0}\in I$ and consider $T^{'}:=T\setminus U_{i_{0}}$. $T^{'}$ is a proper subspace and thus there is a finite subset $I_{F}\subset I$ such that $\left\{U_{i}\right\}_{i\in I_{F}}$ is an open cover for $T^{'}$ but this means that ...

1

A map $f:X\to Y$ is called proper if the preimage of every compact subset is compact. It is called closed if the image of every closed subset is closed. If $X$ is a compact space and $Y$ is a Hausdorff space, then every continuous $f:X\to Y$ is closed and proper. Here are some examples where $f$ is not proper: With $X$ compact: Let $X=[0,1]$ and ...

1

Every second-countable space is separable. We obtain a countable dense subset by choosing one point from each basis set in the countable base. As $X$ is a subset of a second-countable space, it is itself second-countable, since a base is formed by the intersections of the sets in the base of $\Bbb R^d$. Compactness is not needed for this. On the other hand, ...

1

I think I proved the following Lemma Let $X$ be a Hausdorff space and $C \subset X$ have a compact neighbourhood $K$. Then $C$ is a component of $X$ if and only if $C$ is a component of $K$. in this answer. For the present problem this implies a negative answer, since a compact set in a locally compact Hausdorff space has a compact neighbourhood. ...

1

I will use $B(x,r)$ for the open ball centered at $x$ with radius $r$, $\bar B(x,r)$ for the corresponding closed ball and $\overline{B(x,r)}$ for the closure of the open ball in $X$. First, notice that it is sufficient to prove that every closed ball is connected, because of $$B(x,r) = \bigcup_{s<r}\bar B(x,s).$$ (Remember that a union of a family of ...

1

Assume that $X$ is not connected. Then there exists a set $F\subset X$, such that $F, K=X\smallsetminus F$ are non-empty, open and closed, and as they are closed they are compact. This means that $$r=\mathrm{dist}(F,K)>0,$$ and further there exist $x\in F$ and $y\in K$, such that $d(x,y)=r$. In particular, $$\big(F\cup B(x,r)\big)\cap ... 1 For any supposed limit g (continuous function with compact support), take n_0 large enough s.t. for n\ge n_0: {\rm supp}\,f_n\cap{\rm supp}\,g=\emptyset. Now, if n_k\ge n_0:$$ d(f_{n_k},g)^2=\int_{-\infty}^{\infty}|f_{n_k}(t)-g(t)|^2dt= \int_{-\infty}^{\infty}|f_{n_k}(t)|^2dt+\int_{-\infty}^{\infty}|g(t)|^2dt\ge  ...

1

As $A$ and $B$ are closed and bounded subsets of $\mathbb R$, then they are compact. Assume now that $z_n\in A+B$ and $z_n\to z$, we need to show that $z\in A+B$. Then $z_n=x_n+y_n$, with $x_n\in A$ and $y_n\in B$. Since $A$ is compact, then there exists a converging subsequence $x_{n_k}\to x\in A$. But this implies that $$y_{n_k}=z_{n_k}-x_{n_k}\to z-x,$$ ...

1

Since we are in the reals, one can use the fact that a set of reals is compact if and only if it is closed and bounded. Suppose $A\setminus\{a\}$ is compact. Then it is bounded, so $A$ is bounded. And $A$ is the union of the two closed sets $A\setminus\{a\}$ and $\{a\}$. The next two assertions are false. Let $A=[0,1] \cup \{17\}$, and let $a=17$. We ...

1

$\mathbb{R}$ is normal which means for every pair of disjoint closed sets $A, B$ there exists disjoint open sets $U$ and $V$ such that $A \subset U$ and $B \subset V$. Since one point sets are closed in $\mathbb{R}$ and $A - \{a\}$ is closed because it's compact, you can find disjoint open sets $U, V$ that contain $A - \{a\}$ and {$a$} respectively. Then ...

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