I know that the image of a continuous function is bounded, but I'm having trouble when it comes to prove this for vectorial functions. If somebody could help me with a step-to-step proof, that …

You'll find topological properties with indication of whether they are preserved by (various kinds of) continuous maps or not (such as open maps, closed maps, quotient maps, perfect maps, …

A function is "differentiable" if it has a derivative. A function is "continuous" if it has no sudden jumps in it. Until today, I thought these were merely two equivalent definitions of the same c...

Some people like discrete mathematics more than continuous mathematics, and others have a mindset suited more towards continuous mathematics - people just have different taste and …

To find examples and explanations on the internet at the elementary calculus level, try googling the phrase "continuous extension" (or variations of it, such as "extension by continuity") …

For a function to be continuous at point a, the value of f (a) must be the limit of f (x) when x goes to a. In your case there is no f (0), so it cannot be continuous there.

3 This property is unrelated to the completeness of the domain or range, but instead only to the linear nature of the operator. Yes, a linear operator (between normed spaces) is bounded if …

In basic calculus an analysis we end up writing the words "continuous" and "differentiable" nearly as often as we use the term "function", yet, while there are plenty of convenient (and even …

Here you want to refer to the topology of the latter as a normed space, which does not depend on the norm since they are all equivalent in finite dimension. Then the determinant is a polynomial …

Functions in this space are technically only defined up to equality "almost everywhere", so you can change the values of a continuous U function on $ [-1,1]$ arbitrarily on a set of measure …