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8

He means that $X$ and $X'$ are two topological spaces with the same underlying set. $X$ carries the topology $\mathcal T$ and $X'$ the topology $\mathcal T'$. I would have formulated it in the following way: Let $\mathcal T$ and $\mathcal T'$ be topologies on a set $X$ and consider the function \begin{align*} i\colon (X,\mathcal T')&\to (X,\mathcal ...

8

As mentioned by John in the comments, in Hausdorff (T2) spaces, compact subsets are always closed. Suppose $X$ is Hausdorff, and $K \subseteq X$ is compact. To show that $K$ is closed, it suffices to show that $X \setminus K$ is open; that is, show that each $x \in X \setminus K$ has an open neighbourhood $U$ with $U \subseteq X \setminus K$. So given ...

7

The zero set of a polynomial $p\in \Bbb{C}[x,y]$ is unbounded (FTA), but the torus is compact, so that doesn't work. OTOH you can get many toruses as algebraic varieties in the projective space $\Bbb{C}P^2$. Look up Elliptic curves. These can also be described as sets of solutions of an equation of the form $$y^2=x^3+Ax+B$$ together with a point at ...

6

Look at the following diagram (created using http://Presheaf.com. I didn't know how to do the \Bbb-letters there) Since both paths simply identify $(\Bbb N×\Bbb Q×\{0\})\cup(\Bbb N×\{0\}×\Bbb Q)∪(\{0\}×\Bbb Q×\Bbb Q)$ to a point, we have induced set maps $s$ and $\tilde s$, where the later is a bijection. Since the lower left arrow is a quotient map, ...

5

Let $O : \mathbf{Haus}^\mathrm{op} \to \mathbf{Set}$ be the functor that sends a Hausdorff space $X$ to the set of open subsets of $X$. Then $O (1)$ has two elements, so any representation of $O$ must be a two-point space. But the only two-point Hausdorff space is discrete – which clearly won't work.

5

This only really works in the specific case of distinguishing $1$ dimensional and $n$ dimensional Eulcidean space. For higher dimensions you will need a more sophisticated approach. Normally this includes the use of techniques from algebraic topology. See this page for one approach to solving this problem. To comment specifically on why your approach ...

5

The answer is no. For example, $f:[0,1)\rightarrow S^1, t\mapsto e^{2\pi i t}$ is continuous and bijective, however the inverse is not continuous (and this is why this works). Set $A=\{0\}$ and $B=[1/2,1)$. They are closed in $[0,1)$ and disjoint, but $f(A)=\{1\}$ and also $1\in \overline{f(B)}$, because $e^{2\pi i t}\rightarrow 1$ as $t\rightarrow 1-$. ...

4

Here you only need to care about the topology coming from the neighborhood basis. Let $B$ be a topological space. Assuming we can find local trivializations $$f_{U}:U_{E}\cong U\times \mathbb{R}^{n}$$ such that on $U_{E}$ this is given by product topology of $U\times \mathbb{R}^{n}$. Then you "glue" two such neighborhoods together using ...

4

The comb space works although any point on the 'base line' can be deformation retracted to so depending on the interpretation of your question, this might not quite fit your criteria. Instead, we can take a space which is morally the comb space but where we quotient out by the subset of points which can be deformation retracted onto. I believe (though I have ...

4

Let's try to show that $X \setminus S_n(U)$ is open (note that $n$ is fixed from now on): let $p$ be a point such that $p \notin S_n(U)$, so $B(p, \frac{1}{n})$ is not a subset of $U$. This means that for some $q$ with $d(p,q) < \frac{1}{n}$, we have that $q \notin U$. So $q$ witnesses that $p$ is outside $S_n(U)$. The idea is that a small enough open ...

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 ...

3

The crucial fact here is that the fundamental group of a topological group must be abelian. As you point out in the comment, $\pi_{1}(M_{f})$ contains $\pi_{1}(X)$ as a subgroup and hence if $\pi_{1}(X)$ is not abelian, $\pi_{1}(M_{f})$ is neither and so $M_{f}$ cannot be a topological group. The only closed surfaces with abelian fundamental group are the ...

3

The three conditions that must be satisfied by any metric are: $$d(x, y) \geq 0,\; \text{ and }\,d(x, y) = 0 \iff x = y$$ $$d(x, y) = d(y, x)$$ $$d(x, y) + d(y, z) \geq d(x, z), \quad \forall x, y, z \in \mathbb R$$ Your task is to show whether (or not) $d(x, y) = \arctan|x - y|$ satisfies each and every property above. Added: Is $\arctan|x - y|$ ...

3

The product $X\times Y$ is a topological group containing both $X$ and $Y$ as subgroups (as $X\times\{1\}$ and $\{1\}\times Y$). More generally, if $B$ is a topological group and $f:X\to B$ and $g:Y\to B$ are continuous group homomorphisms, then we can construct the fibre product $$X_f\times_g Y = \{(x,y)\in X\times Y: f(x)=g(y)\}$$ which is a closed ...

3

Note that if $(X,d)$ is metric space, then $d'=d/(1+d)$ generates same topology of $(X,d)$. So we only prove this proposition: Let $(X_n,d_n)$ be a sequence of metric spaces, and $d_n(x,y)\le 1$ for all $n$ and $x,y\in X_n$, then $d((x_n),(y_n))=\sum_n 2^{-n} d_n(x_n,y_n)$ generates the product topology of $X=\prod_n X_n$. At first, we prove that for ...

3

$\Rightarrow\quad$ by contradiction If $p\in S$ the result is clear. Now assume $p\not\in S$ and that all ball $B$ centred at $p$ doesn't contain any point of $S$ then $B^c$ is closed containing $S$ and not $p$ so their intersection doesn't contain $p$ as well. Contradiction. $\Leftarrow\quad$ by contraposition Assume $p$ isn't in the closure of $S$ then ...

3

It's not at all clear to me that the two questions in Question 1 are asking the same thing, but in any case the answer to the first one is "definitely not". For example, think about a flat family of elliptic curves degenerating to a nodal rational curve. The general fibre is topologically a torus, so has $H_1 \simeq \mathbf Z \oplus \mathbf Z$, but the ...

2

Q1. No, it's not true that all fibers of a flat map are homotopy equivalent. For instance, the blow-up of $\mathbf P^1 \times \mathbf P^1$ at a point maps to $\mathbf P^1$. This map is flat and all fibers except for one are spheres; however, the remaining fiber is a wedge of two spheres. More generally, in a flat family of curves you can "collapse" a simple ...

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 $\{ X , ... 2 You can do it geometrically. Take a point in your set, and suppose it is in the first quadrant. Drop lines parallel to the axes to the line$x+y=1$, and mark half the distance to the intersection. Then the circle with that radius is entirely contained in the square, because the diagonal is strictly longer that the radius. 2 There are several ways to show it: Let the function$f: \mathbb R^2\to\mathbb R$, with $$f(x_1,x_2)=\lvert x_1\rvert+\lvert x_2\rvert.$$ Clearly$f$is continuous and as$(-\infty,1)$is open in$\mathbb R$, so is its inverse image: $$f^{-1}(-\infty,1)=\big\{(x_1,x_2)\in\mathbb R^2: \lvert x_1\rvert+\lvert x_2\rvert<1 \big\}.$$ 2 I would take a more pedestrian approach. Pick$x \in \overline{A}$. Let$V_i$be a neighbourhood of$[(x,i)]$in$Y$(where$[(x,i)]$denotes the equivalence class of$(x,i)$modulo$R$). By the continuity of the projection, there are open neighbourhoods$U_i$of$(x,i)$in$X\times \{0,1\}$that are mapped into$V_i$. Since$X\times\{i\}$is open in ... 2 The quotient also has to respect the group structure which yours does not. For instance$(0,\frac{1}{4})\sim (1,\frac{3}{4})$but$(0,\frac{1}{4})+(\frac{1}{2},0)=(\frac{1}{2},\frac{1}{4})$and$(1,\frac{3}{4})+(\frac{1}{2},0)=(\frac{1}{2},\frac{3}{4})\nsim(\frac{1}{2},\frac{1}{4})$. So you do not have a well defined group structure on the quotient. To ... 2 Identify$S^2 \cong \mathbb{C}P^1$. Then act on$S^2 \times \ldots \times S^2$by$S_n$. The quotient space is$\mathbb{C}P^n$and the projection map is the branched cover in question. In my case n=2 and the diagonal sphere is the the fixed set under the involution of$S_2$, so it is the branching locus. 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

This isn't really an answer, but it's too long for a comment. In any case, I think the answer could depend on your formulation. I can see two ways to interpret your question. Suppose $X$ and $Y$ are topological groups, homeomorphic as topological spaces. Are they necessarily isomorphic as abstract groups? topological groups? Obviously 2 implies 1. I ...

2

I will restrict myself only to Lie groups, since the world of groups outside of this class is way too large and, I do not think, there is a good answer in this context. I will also restrict to simply-connected Lie groups so that the answer is reasonably neat. (This is not a very good reason, but this will keep my answer reasonably brief, I did not think ...

2

I'll stick with your question 2. By definition a fiber bundle $E\to X$ is locally homeomorphic to $U\times F$ where $U$ is open in $X$ and $F$ is the fiber. So in fact if $X$ is connected all the fibers of $E$ are homeomorphic. On the other hand if $X$ has distinct connected components the fibers over each one can differ arbitrarily, so that there's no ...

2

Let $(O_i)_{i\in I}$ be any open cover of $T$. Since $T$ is regular, one can find an open cover $(V_j)_{j\in J}$ such that, for each $j\in J$, the closure $\overline V_j$ of $V_j$ is contained in $O_{i(j)}$, for some $i(j)\in I$. Next, by assumption on $T$, one can find $j_1, \dots ,j_N\in J$ such that $V_{j_1}\cup\dots \cup V_{j_N}$ is dense in $T$. Then ...

2

Your proof is correct and in my opinion is the simplest way to prove this fact. A more geometric approach is given by Hausdorff-Kuratowski convergence. The compactness theorem for such convergence says in particular that any sequence of compact sets (i.e. your circles) which are all contained in the same compact sets (the bounding square) converge, up to a ...

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