# Tag Info

7

Since the function $f(x,y) = x^2 + y^2 -2x + 4y - 11$ is continuous, it follows that $f^{-1}(\{0\})$ is closed. Since $\lim_{\|(x,y)\| \to \infty} f(x,y) = \infty$, we see that the set must be bounded. To see why this implies that the set is bounded, note that we can find some $R$ such that if $\|(x,y)\| > R$, then $f(x,y) > 1$. Hence the set ...

6

Hint $(x-1)^2+(y+2)^2=16$ is the set so it is a circle and bounded and closed

6

Every sequence needs to have a converging subsequence but not every subsequence needs to converge. In your case obviously $a_{2k}$ converges which is sufficient to see that the sequence is no counter example.

5

Let $f:G\to H$ be an injective Lie group morphism with $H$ compact. Since $H$ has a faithful finite dimensional representation, it suffices to compose with $f$ to obtain one for $G$. This shows that any Lie group $G$ which does not have faithful finite dimensional representations provides an example of what you want.

5

For $n>1$ the space $\Bbb R^n$ has a one-point compactification, but it does not have a $k$-point compactification for any integer $k>1$; see this PDF. $\Bbb R$ itself has both a one-point compactification, which is homeomorphic to the circle $S^1$, and a two-point compactification, which is homeomorphic to $[0,1]$. You might also want to look at the ...

5

The first is easy: $\mathbb{Q}$ is a Hausdorff (metrisable) topological group, that is not locally compact. Or we can use $\mathbb{Z}^\mathbb{N}$ in the product topology (which is a topological group that is (as a topological space) homeomorphic to the irrationals, another non-locally compact space). Or, as the comment suggested, any infinite-dimensional ...

5

Because of your question I assume that the definition of compact that you use does not require the space to be Hausdorff. The answer then depends on your definition of locally compact: If you require every point $x \in X$ to have some compact neighbourhood (weaker condition) then this is always true, because $X$ itself is a compact neighbourhood of $x$. If ...

4

Let us say that a continuous map $f : X \to S$ is universally closed if, for every continuous map $g : Y \to S$, the projection $\{ (x, y) \in X \times Y : f (x) = g (y) \} \to Y$ is closed. The characterisation of compactness you allude to is this: A topological space $X$ is compact if and only if the unique map $X \to 1$ is universally closed. It is ...

4

Let be $X$ any space and $f$ constant...

3

Hint: Let $C \subset \mathbb R^2$ be the set of $x = (x_1, x_2)$ that satisfy $|x_1-2| + |x_2-1| \leq 2$. Then your set is $C \times \mathbb R$. Does that look compact under the usual topology?

3

In a metric space $(X,d)$, the function $d:X\times X\to \Bbb R$ is notoriously continuous. Given a fixed non-empty subset $F\subseteq X$, the function \begin{align}&d(\bullet, F):X\to \Bbb R\\ &d(x,F)=\inf\,\{d(x,y)\,:\, y\in F\}\end{align} is continuous as well. By definition $d(E,F)=\inf\,\{x\in E\, :\, d(x,F)\}$, which, in your case, is actually a ...

3

Some extended HINTS: The members of $\mathscr{T}$ are precisely the sets $f^{-1}[U]$ such that $U\in\mathscr{T}_{Eucl}$. Check that if $\mathscr{B}_{Eucl}$ is a base for $\mathscr{T}_{Eucl}$, then $\mathscr{B}=\left\{f^{-1}[B]:B\in\mathscr{B}_{Eucl}\right\}$ is a base for $\mathscr{R}$. Now take for $\mathscr{B}_{Eucl}$ the set of open intervals in ...

3

Alternative idea: $$\bigcap_{n=1}^\infty E_n = \emptyset\implies \bigcup_{n=1}^\infty(X\setminus E_n) = X$$

2

As you seen, it is a circle with certain center $x_0=(1,-2)$ and radio $r=4$. For the boundness, show that, every point in your set has lenght $<7\sqrt{2}$. For the closeness, show that the complement is open (you only have to follow the geometry of your set). Can you continue? Edit $\begin{eqnarray} ... 2 For the second part, take a bounded subset$B$of$C[a,b]$. We can find a constant$R$such that$|v(x)|\leqslant R$for each$x\in [a,b]$and each$v\in B$. To prove that$M(B)$is equi-continuous and bounded, use the fact that$g$is uniformly continuous (and bounded) on$[a,b]\times [-R,R]$. 2 Let us assume$X$and$Y$are Hausdorff (otherwise, I'm not sure what "the one point compactification" is supposed to mean). Then$X\wedge Y$is compact Hausdorff, being the quotient of the compact Hausdorff space$X\times Y$by a closed equivalence relation (the equivalence relation is closed because it is the union of the diagonal in$(X\times Y)^2$and ... 2 This is true trivially. A space is locally compact if every point has a compact neighborhood. If the space itself is compact, then it is a compact neighborhood of every point. 2 If$X$is a locally compact Hausdorff space, one says that a function$f\colon X \to \mathbb{R}$"vanishes at infinity" if for every$\varepsilon > 0$there is a compact$K_{\varepsilon} \subset X$such that$\lvert f(x)\rvert < \varepsilon$for all$x\in X\setminus K_{\varepsilon}$. If$X^{\ast} = X \cup \{\infty\}$is the one-point compactification ... 2 From $$f(x,y,z):=x^2+y^2+z^2+xy+yz+zx={1\over2}\bigl((x+y+z)^2+x^2+y^2+z^2\bigr)$$ it follows that $$x^2+y^2+z^2\leq 2f(x,y,z)=2\qquad\forall\ (x,y,z)\in M\ .$$ This shows that$M$is (not only closed, but) also bounded, hence compact. 2 For$b \in C$let$A_b := \{ f(a) : a < b \}$(so that$U_b = \bigcup A_b$). I'll back up a bit, and give a fairly full justification for why$A_b$cannot cover$X$. If$A_b$covers$X$(so that$U_b = X$), since$A_b$is a subfamily of$A$it must be that$A_b$does not have a finite subcover (otherwise$A$would). But the choice of the well-ordered ... 2 Hint: This is an alternative to your approach that's more straightforward. For all$x \in C[0,1]$(hence, for all$x \in K$) $$|f(x)| \leq \|x\|_\infty\|y\|_\infty.$$ (Why? Show this if you don't have a theorem that proves this already.) Hence $$\|x_n\|_\infty \to 0\text{ as }n \to \infty \implies |f(x_n)| \to 0\text{ as }n\to\infty,$$ as well. 2 It's not silly question. The part you didn't understand is: "has a subsequence"= "has at least one" - that means it is enough to have just one subsequence that converges in$A$, for example$a_{2k}$but all others don't have to converge, for example sequence$a_k$as subsequence itself. Maybe counterexample will help you to understand more: You can use ... 2 The simplest example is an infinite set$X$equipped with the cofinite topology. This satisfies the first three conditions, but the whole space$X$is an irreducible closed set without a generic point. This also provides a good example of soberification: we can make$X$into a sober space by adding a single dense point. Note: I am taking "compact" to ... 2 There is no such exercise on page 36, in either the first or second edition of my book. You must be using one of the pirated draft versions of the first edition, which somebody posted illegally on the internet. Those are full of mistakes and come with no guarantees. 2 The image of$(-1,1]$by$x \to x^2$is compact. 1 Every open set in$\mathbb{R}^k$is a increasing union of closed sets. Namely, if the open set is$A$, take$C_n:=\{x \mid d(x,A^c) \geq \frac{1}{n}\}.$Consider now the sequence$K_n=[-n,n]^k$. We then have that the sequence$K_n \cap C_n$is what you need. Note that the argument holds for any$\sigma$-compact metric space, and the fact that an open set ... 1 Yes: a compact metric space is second countable, and second countability is hereditary and implies separability. 1 Hint: Let$x_n$be a sequence of$X$which converges towards$x$, since$K$is compact we can extract$x_{l(n)}$such that$f(x_{l(n)})$converges towards$y$. Since$(x_{l(n)},f(x_{l(n)}))$is in the graph and the graph is closed,$(x,y)$is in the graph. This implies that$f(x)=y$and$f(y)$is adherent to every convergent subsequence of$f(x_n)$thus$f$... 1 Hint: Say that you have a sequence$x_n\to x$such that$x_n(0) \in [-3,4]$and$|x_n(t)-x_n(s)| \leq d |t^2-s^2|$for all$t$and$s$(you mean in$[0,1]$), for each$n$. What happens in these two properties if$n\to\infty$? 1 Suppose$\;\{x_n\}_{n=1}^\infty\subset K\;$is such that$\;x_n\xrightarrow[n\to\infty]{}x\;$. We must show that$\;x\in K\;$. By continuity: $$x(0)=\lim_{n\to\infty}x_n(0)\in[-3,4]\;,\;\;\text{as this last is a closed interval}$$ Also, again by continuity, for all$\;t,s\in[0,1]\;\$ , we have ...

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