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Continuum theory, point-set topology, spaces with algebraic structure, foundations, dimension theory, local and global properties.

11 votes

When can boundedness be characterized topologically in Metric spaces?

It is an old result of Klee saying that the infinite-dimensional Hilbert space is homeomorphic with both its unit sphere and its closed unit ball. See e.g. http://www.ams.org/journals/bull/1961-67-03/ …
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4 votes

Injective maps $\mathbb{R}^{n} \to \mathbb{R}^{m}$

Alternatively, you may use the Borsuk-Ulam antipodal theorem, in order to prove that such a map cannot be one-to-one.
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  • 4,060
41 votes

Is a topology determined by its convergent sequences?

Just another example. Consider the Banach space $\ell^{1}\left(\Gamma\right)$ , $\Gamma$ being an infinite set. Then the weak topology and the norm topology have the same convergent sequences (Schur' …
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12 votes
2 answers
2k views

Topological Rings

Is it true that, if S is a subring of a separable topological Noetherian ring R, then S is separable, too ?
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6 votes

Does Cauchy continuity imply uniform continuity? [No.]

Well, here is [most probably another] Wikipedia page: http://en.wikipedia.org/wiki/Cauchy-continuous_function. HTH. Alternatively, you may use the function $f:\mathbb{R\rightarrow\mathbb{R}}$ , expre …
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  • 4,060
3 votes

Cone in a metric space

First of all, the notion of $cone$ is a purely algebraic stuff, and not a metrical one. The $cone$ is naturally defined in the framework of $linear$ spaces, and not of Banach spaces. One can introduce …
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21 votes

solving $f(f(x))=g(x)$

Q2) has a negative answer. Namely, if, e.g., $g(x)=-x$ for all $x\in\mathbb{R}$, then there is no continuous $f:\mathbb{R\rightarrow\mathbb{R}}$ such that $f\circ f=g$. As to Q3, see, e.g., Theor …
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1 vote

Extending Continuous Sublinear maps on dense subsets of a Banach space

To 1) and 2). Let $X^{'}=Y=c_{0}$ , and let $X=c_{00}$. Take some $p\in X^{'}\smallsetminus X$, and define $T:X\rightarrow Y$ by $Tx:=\left\Vert x\right\Vert \cdot\left(\sin\left(\left\Vert x-p\right\ …
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12 votes
Accepted

Sequential topological vector spaces

The space of tempered distributions is sequential (for its usual strong topology). See, e.g., Dudley, and the references therein.
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2 votes
1 answer
507 views

Are the C(S^n, S^n)'s homeomorphic ?

Let m, n > 1. Is it true that C(S^m, S^m), and C(S^n, S^n) are homeomorphic ? [both endowed with the sup metric (or equivalently the compact-open topology)] Generally, C(S^n, S^n), with n >= 1, is a …
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