Intereting Posts

Explicit bijection between ordered trees with $n+1$ vertices and binary trees with $n+1$ leaves
Conformal mapping of a doubly connected domain onto an annulus
Intuition behind the Axiom of Choice
Composite functions and one to one
Real Analysis Methodologies to show $\gamma =2\int_0^\infty \frac{\cos(x^2)-\cos(x)}{x}\,dx$
Trigonometric Equation : $\sin 96^\circ \sin 12^\circ \sin x = \sin 18^\circ \sin 42^\circ \sin (12^\circ -x)$
if $\Omega=\{1,2,3,\cdots \}$ then $S_{\Omega}$ is an infinite group
Gerrymandering/Optimization of electoral districts for one particular party
A problem on sinusoids
What is the advantage of measuring an angle in radian(s)?
Proving that $\mathbb{Q}$ is neither open or closed in $\mathbb{R}$
What does the symbol $\subset\subset$ mean?
solutions of $a^2+b^2=c^2$
Is every skew-adjoint matrix a commutator of two self-adjoint matrices
Which one result in mathematics has surprised you the most?

Let $S$ be the set $S = \{(e^{-x}\cos (x), e^{-x}\sin(x)) : x\geq 0\} \cup \{(x,0):0\leq x \leq 1\}$.

Is $S$ compact? I know I have to show that if $S$ is closed and bounded, then it is compact. Since the set contains $x=0$ though and $x$ is never less than $0$, is it clearly compact? Or is there something more to show?

- Closed set in $\ell^1$
- shortest path to Tychonoff?
- Tychonoff theorem (1/2)
- Exercise on compact $G_\delta$ sets
- Proving the inverse of a continuous function is also continuous
- A relative compactness criterion in $\ell^p$

- Uncountable family of uncountable compact subsets of $\mathbb{R}$
- Continuity of $f \cdot g$ and $f/g$ on standard topology.
- Why does “separable” imply the “countable chain condition”?
- Pseudocompactness in the $m$-topology in $C(X)$
- Showing that $$ is compact
- Any idea about N-topological spaces?
- show that a subset of $\mathbb{R}$ is compact iff it is closed and bounded
- Locally connected and compact Hausdorff space invariant of continuous mappings
- Why this two spaces do not homeomorphic?
- Prove that if $X$ and it's closure $\overline X$ are connected and if $X\subset Y \subset \overline X$, show that Y is also connected.

The only limit point of the first set $S_1$ but not in $S_1$ is $(0,0)$. Since for any $p=(u,v)\neq (0,0)$ and $p$ is not an element of the first set $S_1$, let $r^2=u^2+v^2$, we can find sufficient large $x$, s.t $e^{-x}<r$. Then we can find a neighborhood of $p$, which doesn’t intersect $S_1$. Hence $p$ is not a limit point.

Since $(0,0)$ is included in the second set $S_2=\{(x,0):0\leq x \leq 1\}$ and $S_2$ is closed, the union $S=(S_1 \cup \{(0,0)\}) \cup S_1$ should be closed. And it’s bounded since it’s contained in the unit square. Hence it’s compact.

- How I could define a inner product in the characters in $SL(2, \mathbb R)$
- The boundary of an $n$-manifold is an $n-1$-manifold
- Is there any easy way to understand the definition of Gaussian Curvature?
- Visualizing the four subspaces of a matrix
- Three points on sides of equilateral triangle
- A non-square matrix with orthonormal columns
- algebraic sum of a graph of continuous function and itself Borel or measurable?
- How is $L_{2}$ Minkowski norm different from $L^{2}$ norm?
- Pushforward of Lie Bracket
- Is there a “continuous product”?
- $\sum\limits_{k\in\mathbb{Z}}$ versus $\sum\limits_{k=-\infty}^{\infty}$
- Complex numbers – roots of unity
- A trigonometric identity for special angles
- Proof of equicontinuous and pointwise bounded implies compact
- Does $f(X \setminus A)\subseteq Y\setminus f(A), \forall A\subseteq X$ imply $f$ is injective ?