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Definitely, the one I like the most is the proof via ultrafilters. You only have to state the compactness of a topological space in terms of ultrafilters, which is easily obtained by the definition vi …

The first relevant fact about $f$ is that it is a proper map. In such a situation the topological (Brouwer) degree of $f$ is well-defined, and by the product rule $\operatorname{deg}(T)= \operatorname …

The existence of such functions being established by now, there remains a curiosity for the concrete situations that may generate them. Here is one, which is also interesting as an example of a $C^\in …

They are not ring isomorphic, because e.g. $R$ has the following property of a ring $X$, and $S$ does not: There is a non zero element $u \in X$ such that for any invertible $f\in X$ either $u …

Yes. For any compact subset $K\subset X$ consider a uniformly continuous extension of $f_{|K}$ (there is such an extension even with the same subadditive modulus of continuity, see here). This shows t …

Recall the definition of the Peano square-filling curve $f:[0,1]\to[0,1]^2$, which is given in terms infinite ternary strings. If $a\in [0,1]$ has a base $3$ representation of the form $0,a_1a_2a_3\d …

No: if $X$ is non compact, it is a proper and dense subset, thus not closed, in its Stone–Čech compactification. [edit] This is ok e.g. if X is $T_{3.5}$, as Bruno observes, otherwise $X\to\beta X$ …

Not an answer to the question, but a hopefully related observation to complete the proof and explain why the result is not obvious, which is may be of interest for your class. I would mention that a …

Actually real open intervals with rational left end-point and irrational right end-point are a base with that property.

As I said in the comment above, I think that the set of points $t\in[0,1]$ where a square-filling curve with strictly increasing area defines a convex $f([0,t])$, is a nowhere dense closed set contain …

Consider the topology on $\mathbb{R}^2$ generated by subsets that are open in some line from the origin. This topological space is connected, has the cardinality of continuum, and has $2^c$ open subse …

As to the compact-open topology of $C(X,Y)$, it is metrizable if and only if $Y$ is metrizable, and $X$ is hemicompact.

Here is a counterexample for $K:=[0,1]$. There exists a Borel set $B\subset K$ that meets every non-empty open set $A\subset K$ in a subset of it of positive, not full Lebesgue measure: $0<|A\cap B|<| …

A bump function on a Banach space $X$ is a non-zero function with bounded support. Existence of a bump function of a given smooth regularity has, of course, strong immediate consequences --by translat …

To summarize the situation. Let $(X,\mathcal{F})$ a measurable space. The space $M(X,\mathcal{F})$ of all real-valued signed measures on $(X,\mathcal{F})$ is a Banach space wrto the total variation no …