I am looking for a direct proof (not by contradiction) that a compact metric space is sequentially compact, ie constructing a converging subsequence from any sequence. Thanks

Let $(W, \leq)$ be a linear order, and let $f : [0, 1] \rightarrow W$ be a continuous function (where [0, 1] has the usual topology and W has its order topology). Show that the range of f is convex. Let $Y \subseteq X$. The subspace $Y$ is a convex set if for each pair […]

If a set $X$ in a topological space $T$ has the property that for all sequences $x_n \in X, x_n \to x \implies x\in X$ must X be closed? I know this is true for metric spaces but is it true for a general topological space. Here I am defining $X$ is closed iff $T\setminus […]

If $U$ is an open subset of $\mathbb{R}^2$, is it true that $H_2(U)=0$ and what can we say about $\pi_1(U)$? (For example, can we show that $\pi_1$ isn’t perfect, e.g. $\pi_1(U)\neq 0\Rightarrow H_1(U)\neq 0$?) For $\pi_1$, I’m aware that there is an answer here: fundamental groups of open subsets of the plane, but I’d rather […]

This is my first post on the mathematics stack exchange site. If I am posting this question on the wrong site, I apologize and please let me know so I can delete it. I am a programmer, and one thing we have to worry about is optimization. This is why my question probably isn’t as […]

There are several proofs I have seen of this, but they all seem to use choice subtely at some point. Is there any way to prove this without choice, or is it possibly unproveable?

This originally comes from $f_1(x,y)=\frac{x}{y}$, where $X=\mathbb{R}^{n}, Y=\mathbb{R}^m, x \in X, f: X \rightarrow Y, x \neq 0, f(x) \neq 0$ $$\limsup_{(h_x,h_y)\to(0,0)} \frac{\left|\frac{x+h_x}{y+h_y}-\frac{x}y\right|} {\sqrt{{h_x}^2+{h_y}^2}}$$ If I understand it correctly, $x$ and $y$ can be anything but zero, and $h_x,h_y$ go towards zero. Moreover, both numerator and denominator cannot be negative. But since $x,y$ could be […]

My question is: Show that the lexicographic order topology for $\mathbb{N}\times \mathbb{N}$ is not the discrete I have been thinking on the fact that on the discrete topology all singleton sets are open. If can I find one singleton not open the proof is done? For example $\{(0,0)\}$ is not open because it has not […]

The problen can be foun in p.71 of Topology and Geometry. I state it below for convenience. Problem: Consider the half open real line $X=[0,\infty)$. Define a functional structure $F_{1}$ by taking $f\in F_{1}(U) \Leftrightarrow f(x)=g(x^{2})$ for some $C^{\infty}$ function $g$ on $\{x| x\in U\text{ or} -\negmedspace x\in U\}$. Define another functional structure $F_{2}$ by […]

Let $(X,\tau)$ be a topological space. Suppose $dc(X)=\kappa$ and let $D\subset_{dense} X$ be a dense subset of $X$ of cardinality $\kappa$. Is it true that $X\setminus D$ has density character $\kappa$, as a subspace of $X$ with the restricted topology?

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